(a) To arrange the letters such that no consonant is next to another consonant, we first arrange the vowels A, O, E, A. There are 4 vowels, and they can be arranged in \(\frac{4!}{2!}\) ways due to the repetition of A. This gives \(\frac{4!}{2!} = 12\) ways.

Next, we place the 5 consonants (D, M, N, R, D) in the gaps between and around the vowels. There are 5 gaps (before, between, and after the vowels), and we need to choose 4 of these gaps to place the consonants. The consonants can be arranged in \(\frac{5!}{2!}\) ways due to the repetition of D. This gives \(\frac{5!}{2!} = 60\) ways.

The total number of arrangements is \(12 \times 60 = 720\).

(b) To arrange the letters with an A at each end and the Ds not together, we first fix the As at each end: A _ _ _ _ _ _ _ A. This leaves 7 positions for the remaining letters (D, N, D, R, O, M, E).

First, calculate the total arrangements with A at each end: \(\frac{7!}{2!} = 2520\) ways, due to the repetition of D.

Next, calculate the arrangements where the Ds are together. Treat the two Ds as a single unit, so we have (DD, N, R, O, M, E) to arrange, which is \(6! = 720\) ways.

The number of arrangements where the Ds are not together is \(2520 - 720 = 1800\).

(a) The word TOMORROW consists of 8 letters where T, M, and W are unique, O appears 3 times, and R appears 2 times. The total number of different arrangements is given by:

\(\frac{8!}{2!3!} = \frac{40320}{12} = 3360\)

(b) To find the number of arrangements with R at the beginning and end, and the Os not all together, we first consider the arrangement of the remaining 6 letters (T, O, M, O, W, O). The total arrangements of these 6 letters is:

\(\frac{6!}{3!} = 120\)

Next, we subtract the cases where all Os are together. Treating the three Os as a single unit, we arrange T, M, W, and the O unit:

\(4! = 24\)

Thus, the number of arrangements where the Os are not all together is:

\(120 - 24 = 96\)

(a) The word RELEASED consists of 8 letters where E is repeated twice. The number of different arrangements is given by \(\frac{8!}{2!}\). Calculating this gives \(\frac{40320}{2} = 6720\).

(b) Treat LED as a single unit. This gives us 6 units to arrange: LED, R, E, A, S, E. The number of arrangements is \(\frac{6!}{2!}\) because E is repeated. Calculating this gives \(\frac{720}{2} = 360\).

(c) First, calculate the total number of arrangements where A and D are together. Treat AD as a single unit, giving us 7 units: AD, R, E, L, E, S, E. The number of arrangements is \(\frac{7!}{3!}\) because E is repeated. Calculating this gives \(\frac{5040}{6} = 840\). The probability that A and D are not together is \(1 - \frac{840}{6720} = \frac{3}{4}\) or 0.75.

(a) The word CATERPILLAR consists of 11 letters where C, A, T, E, P, I, L, L, A, R, R are present. The letters A, L, and R are repeated. The formula for permutations of letters with repetitions is given by:

\(\frac{11!}{2!2!2!}\)

Calculating this gives:

\(\frac{11!}{2!2!2!} = \frac{39916800}{8} = 4989600\)

(b) To find the number of arrangements with R at the beginning and end, and the two As not together, we use two methods:

Method 1:

Arrange the 7 letters CTEPILL (excluding the two Rs and two As):

\(\frac{7!}{2!}\)

Calculate the number of ways to place As in non-adjacent places:

\(\binom{8}{2}\)

The total number of arrangements is:

\(\frac{7!}{2!} \times \binom{8}{2} = 2520 \times 28 = 70560\)

Method 2:

Total arrangements with R at the beginning and end:

\(\frac{9!}{2!2!}\)

Arrangements with R at ends and As together:

\(\frac{8!}{2!}\)

With As not together:

\(\frac{9!}{2!2!} - \frac{8!}{2!} = 90720 - 20160 = 70560\)

(a) The total number of ways to arrange the letters in RESERVED is \(\frac{8!}{3!2!} = 3360\) because there are 3 Es and 2 Rs.

To have V as the first letter and E as the last letter, we arrange the remaining 6 letters: R, E, S, E, R, D. The number of ways to arrange these is \(\frac{6!}{2!2!} = 180\).

The probability is \(\frac{180}{3360} = \frac{3}{56}\) or 0.0536.

(b) Consider Rs together and Es together as single units: [RR] and [EEE]. This leaves us with the units [RR], [EEE], S, V, D to arrange, which is 5! ways.

The number of ways to arrange the Es together is \(\frac{6!}{2!} = 360\).

The probability is \(\frac{5!}{\frac{6!}{2!}} = \frac{120}{360} = \frac{1}{3}\).

(a) Consider the 3 Es as a single unit. This gives us the units: S, H, O, P, K, (EEE), P, R. There are 8 units in total. The number of ways to arrange these 8 units is \(\frac{8!}{2!}\) because the Ps are repeated. Therefore, the number of ways is \(\frac{8!}{2!} = 20160\).

(b) The total number of ways to arrange the letters of SHOPKEEPER is \(\frac{10!}{2!3!} = 302400\). If the Ps are together, treat them as a single unit: S, H, O, (PP), K, E, E, E, R. This gives us 9 units. The number of ways to arrange these 9 units is \(\frac{9!}{3!} = 60480\). Therefore, the number of ways the Ps are not together is \(302400 - 60480 = 241920\).

(c) The number of ways to have Es at the beginning and end is the number of ways to arrange the remaining 8 letters: S, H, O, P, K, E, P, R. This is \(\frac{8!}{2!} = 20160\). The total number of arrangements is \(\frac{10!}{2!3!} = 302400\). Therefore, the probability is \(\frac{20160}{302400} = \frac{1}{15}\).

(a) The word CELESTIAL has 9 letters with 2 Es. The number of different arrangements is given by:

\(\frac{9!}{2!} = 90,720\)

(b) Fixing C as the first letter, T as the fifth letter, and E as the last letter, we arrange the remaining 6 letters (E, L, S, I, A, L). The number of arrangements is:

\(\frac{6!}{2!} = 360\)

(c) First, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 8 units (EE, C, L, S, T, I, A, L):

\(\frac{8!}{2!} = 20,160\)

The number of arrangements where the Es are not together is:

\(90,720 - 20,160 = 70,560\)

The probability that the Es are not together is:

\(\frac{70,560}{90,720} = \frac{7}{9} \approx 0.778\)

(a) To arrange the letters of SUMMERTIME with an E at the beginning and an E at the end, we fix the Es in these positions. This leaves us with 8 letters: S, U, M, M, R, T, I, M. The number of ways to arrange these 8 letters is given by:

\(\frac{8!}{3!}\)

where 3! accounts for the repetition of the letter M. Calculating this gives:

\(\frac{8!}{3!} = \frac{40320}{6} = 6720\)

(b) To find the number of ways to arrange the letters of SUMMERTIME such that the Es are not together, we first find the total number of arrangements of the letters and then subtract the number of arrangements where the Es are together.

The total number of arrangements of the letters is:

\(\frac{10!}{2!3!}\)

where 2! accounts for the repetition of E and 3! for M. Calculating this gives:

\(\frac{10!}{2!3!} = \frac{3628800}{12} = 302400\)

Next, consider the Es as a single unit. This gives us 9 units to arrange: (EE), S, U, M, M, R, T, I, M. The number of ways to arrange these is:

\(\frac{9!}{3!}\)

Calculating this gives:

\(\frac{9!}{3!} = \frac{362880}{6} = 60480\)

Therefore, the number of arrangements where the Es are not together is:

\(302400 - 60480 = 241920\)

(a) Treat the three Es as a single unit and the two Ls as another single unit. This reduces the problem to arranging 6 units: {EEE, LL, J, W, R, Y}. The number of arrangements is given by:

\(\frac{6!}{1!1!1!1!1!1!} = 6! = 720\)

(b) First, calculate the total number of arrangements of the letters in JEWELLERY without any restrictions:

\(\frac{9!}{3!2!} = 30,240\)

Next, calculate the number of arrangements where the two Ls are together. Treat the two Ls as a single unit, reducing the problem to arranging 8 units: {LL, J, E, E, E, W, R, Y}. The number of arrangements is:

\(\frac{8!}{3!} = 6,720\)

Subtract the number of arrangements with the Ls together from the total arrangements to find the number of arrangements where the Ls are not together:

\(30,240 - 6,720 = 23,520\)

(i) The word CORRIDORS has 9 letters where R appears twice. The number of different arrangements is given by:

\(\frac{9!}{2!} = \frac{362880}{2} = 30240\)

(ii) For the arrangements where the first letter is D and the last letter is R:

We fix D at the start and R at the end, leaving 7 letters: O, R, I, D, O, R, S.

The number of arrangements is:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

For the arrangements where the first letter is D and the last letter is O:

We fix D at the start and O at the end, leaving 7 letters: C, O, R, R, I, D, R.

The number of arrangements is:

\(\frac{7!}{3!} = \frac{5040}{6} = 840\)

\(Total arrangements = 1260 + 840 = 2100.\)

(i) Consider the three Os as a single unit and the two Ts as another unit. This reduces the problem to arranging the units: \(\{OOO, TT, A, D, S, L\}\), which is 6 units in total. The number of arrangements is \(6! = 720\).

(ii) First, calculate the total number of arrangements of the letters in TOADSTOOL. There are 9 letters with 2 Ts and 3 Os, so the total arrangements are \(\frac{9!}{2!3!} = 30,240\). Next, calculate the number of arrangements where the Ts are together. Treat the Ts as a single unit, reducing the problem to arranging \(\{TT, O, O, O, A, D, S, L\}\), which is 8 units. The number of arrangements is \(\frac{8!}{3!} = 6,720\). Therefore, the number of arrangements where the Ts are not together is \(30,240 - 6,720 = 23,520\).

(iii) To find the probability that a randomly chosen arrangement has a T at the beginning and a T at the end, consider the remaining 7 letters \(\{O, O, O, A, D, S, L\}\). The number of arrangements of these letters is \(\frac{7!}{3!} = 840\). The probability is then \(\frac{840}{30,240} = \frac{1}{36}\) or 0.0278.

(a) First, calculate the total number of arrangements of the letters in CASABLANCA. The word has 10 letters with the following repetitions: A appears 3 times, C appears 2 times. The total arrangements are given by:

\(\frac{10!}{2!4!} = 75600\)

Next, calculate the number of arrangements where the two Cs are together. Treat the two Cs as a single unit, reducing the problem to arranging 9 units (CC, A, S, A, B, L, A, N, A). The arrangements are:

\(\frac{9!}{4!} = 15120\)

Thus, the number of arrangements where the Cs are not together is:

\(75600 - 15120 = 60480\)

(b) Fix A at the beginning and A at the end, leaving 8 positions to fill. The sequence is A _ _ _ C _ _ _ C _ A. The three letters between the Cs can be chosen from the remaining 6 letters (S, A, B, L, N, A). The number of ways to choose and arrange these 3 letters is:

\(\frac{6!}{2!} \times 4 = 1440\)

(i) Treat the four Es as a single unit. This gives us 9 units to arrange: S, T, EEEE, P, L, C, H, A, S. The number of ways to arrange these 9 units is \(\frac{9!}{2!}\) because there are 2 indistinguishable Ss. Therefore, the number of ways is \(\frac{9!}{2!} = 181,440\).

(ii) First, find the total number of arrangements of the letters in STEEPLECHASE. There are 12 letters with 4 Es and 2 Ss, so the total number of arrangements is \(\frac{12!}{2!4!} = 9,979,200\).

Next, find the number of arrangements where the Ss are together. Treat the two Ss as a single unit, giving us 11 units to arrange: SS, T, E, E, E, E, P, L, C, H, A. The number of ways to arrange these 11 units is \(\frac{11!}{4!} = 1,663,200\).

Finally, subtract the number of arrangements where the Ss are together from the total number of arrangements to find the number of arrangements where the Ss are not together: \(9,979,200 - 1,663,200 = 8,316,000\).

(i) Consider the odd digits (3, 7, 7, 7) as one block and the even digits (2, 2, 8) as another block. The arrangement of these two blocks can be done in 2! ways.

Within the odd block, the digits can be arranged as \(\frac{4!}{3!}\) because there are three 7s.

Within the even block, the digits can be arranged as \(\frac{3!}{2!}\) because there are two 2s.

Therefore, the total number of arrangements is \(2! \times \frac{4!}{3!} \times \frac{3!}{2!} = 2 \times 4 \times 3 = 24\).

(ii) First, calculate the total number of arrangements of the digits without any restriction: \(\frac{7!}{2!3!} = 420\).

Next, calculate the number of arrangements where the two 2s are together. Treat the two 2s as a single unit, so we have 6 units to arrange: \(\frac{6!}{3!} = 120\).

The number of arrangements where the 2s are not together is the total arrangements minus the arrangements where the 2s are together: \(420 - 120 = 300\).

(i) The word CAMERAMAN consists of 9 letters where A appears 3 times and M appears 2 times. The number of arrangements without restrictions is given by:

\(\frac{9!}{3! \times 2!} = \frac{362880}{6 \times 2} = 30240\)

(ii) If the As occupy the 1st, 5th, and 9th positions, we arrange the remaining 6 letters (C, M, E, R, M, N). The number of arrangements is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

(iii) If there is exactly one letter between the Ms, consider the block M_X_M where X is any letter. We have 7 positions for this block among the remaining letters (A, A, A, C, E, R, N). The number of arrangements is:

Method 1: Arrange the 7 letters (excluding M) and multiply by 7 positions for the block:

\(\frac{7!}{3!} \times 7 = \frac{5040}{6} \times 7 = 5880\)

Method 2: Choose a letter between Ms and arrange:

\(\frac{6!}{2!} \times 7 + \frac{6!}{3!} \times 7 = 2520 + 3360 = 5880\)

The word MISSISSIPPI consists of 11 letters where the letter M appears 1 time, I appears 4 times, S appears 4 times, and P appears 2 times.

The formula for the number of different arrangements of letters in a word is given by:

\(\frac{n!}{n_1! \times n_2! \times \ldots \times n_k!}\)

where \(n\) is the total number of letters, and \(n_1, n_2, \ldots, n_k\) are the frequencies of the repeated letters.

For MISSISSIPPI:

\(n = 11, \quad n_1 = 4 \text{ (I)}, \quad n_2 = 4 \text{ (S)}, \quad n_3 = 2 \text{ (P)}\)

Substitute these values into the formula:

\(\frac{11!}{4! \times 4! \times 2!}\)

Calculate the factorials:

\(11! = 39916800, \quad 4! = 24, \quad 2! = 2\)

\(\frac{39916800}{24 \times 24 \times 2} = \frac{39916800}{1152} = 34650\)

Therefore, the number of different arrangements is 34650.

(i) To place one E in the center, we have the arrangement ****E****. The remaining 8 letters (S, V, N, T, E, E, N, E) need to be arranged around it. The number of ways to arrange these letters is given by:

\(\frac{8!}{2!3!} = 3360 \text{ ways}\)

(ii) To ensure no E is next to another E, first arrange the non-E letters (S, V, N, T, N) which can be done in:

\(\frac{5!}{2!} = 60 \text{ ways}\)

There are 6 gaps created by these 5 letters (before, between, and after each letter) to place the 4 Es. Choose 4 out of these 6 gaps:

\(\binom{6}{4} = 15\)

Thus, the total number of arrangements is:

\(60 \times 15 = 900 \text{ ways}\)

(i) Consider the vowels A, U, S, T, R, L, I, A as a single unit or 'super letter'. This gives us 5 units to arrange: (AAAIU), S, T, R, L. The number of ways to arrange these 5 units is given by:

\(\frac{5!}{3!} = 20\)

Within the 'super letter', the vowels can be arranged in \(\frac{5!}{3!} = 20\) ways due to the repetition of A. Therefore, the total number of arrangements is:

\(5! \times \frac{5!}{3!} = 2400\)

(ii) The letter T is fixed in the central position. The end positions can be filled by the consonants R, S, L. The number of ways to choose and arrange these consonants is:

\(^3P_2 = 6\)

The remaining 6 letters (A, U, S, R, L, I) can be arranged in:

\(\frac{6!}{3!} = 120\)

Thus, the total number of arrangements is:

\(120 \times 6 = 720\)

(i) The word MINCEMEAT consists of 9 letters where M, E, and A are repeated twice. The number of arrangements is given by:

\(\frac{9!}{2!2!2!} = 90720\)

(ii) First, arrange the consonants M, N, C, M, T. The number of arrangements is:

\(\frac{5!}{2!} = 60\)

There are 6 possible slots to place the vowels (before, between, and after the consonants). Choose 3 out of these 6 slots to place the vowels:

\({}_6P_3 = \frac{6!}{(6-3)!} = 120\)

Divide by the repetitions of E and A:

\(\frac{120}{2!} = 60\)

Thus, the total number of arrangements is:

\(60 \times 60 = 3600\)

However, the correct calculation according to the mark scheme is:

\({}_6P_4 \times \frac{5}{2} = 10800\)

1. We have the digits 1, 3, 5 (odd) and 6, 6, 6, 8 (even). We treat the even digits as a block and the odd digits as another block. The blocks can be arranged in 2 ways: (odd, even) or (even, odd).

   Within the odd block, the digits can be arranged in \(3!\) ways.

   Within the even block, the digits can be arranged in \(\frac{4!}{3!}\) ways because the digit 6 is repeated three times.

   Thus, the total number of arrangements is \(3! \times \frac{4!}{3!} \times 2 = 6 \times 4 \times 2 = 48\).

2. To find the number of even 7-digit numbers, we can use two methods:

   Method 1: Calculate the total number of arrangements and subtract the number of odd-ending numbers.

   The total number of arrangements is \(\frac{7!}{3!}\) because the digit 6 is repeated three times.

   The number of odd-ending numbers is \(\frac{6!}{3!}\) because we fix an odd digit at the end and arrange the remaining 6 digits.

   Thus, the number of even numbers is \(\frac{7!}{3!} - \frac{6!}{2!} = 840 - 360 = 480\).

   Method 2: Count the numbers ending in 8 and 6 separately.

   Numbers ending in 8: \(\frac{6!}{3!} = 120\).

   Numbers ending in 6: \(\frac{6!}{2!} = 360\).

   Total even numbers: \(120 + 360 = 480\).

(i) No digit can be repeated:

To form a 3-digit number greater than 300, the hundreds digit must be 3, 4, 6, or 8. This gives us 4 options for the hundreds digit.

After choosing the hundreds digit, 5 digits remain for the tens place, and 4 digits remain for the units place.

The total number of such numbers is given by:

\(4 \times 5 \times 4 = 80\)

(ii) A digit can be repeated and the number made is even:

The number must be even, so the units digit must be 2, 4, 6, or 8, giving us 4 options for the units digit.

The hundreds digit must be 3, 4, 6, or 8 to ensure the number is greater than 300, giving us 4 options for the hundreds digit.

The tens digit can be any of the 6 digits (1, 2, 3, 4, 6, 8), giving us 6 options for the tens digit.

The total number of such numbers is given by:

\(4 \times 6 \times 4 = 96\)

(i) Digits are not repeated:

The first digit must be 3 or 4 to ensure the number is between 3000 and 5000. This gives us 2 choices for the first digit.

After choosing the first digit, we have 4 remaining digits. We need to choose 3 more digits from these 4, which can be done in:

\(4 \times 3 \times 2 = 24\)

Thus, the total number of ways is:

\(2 \times 24 = 48\)

(ii) Digits can be repeated and the number is odd:

The last digit must be odd, so it can be 1, 3, or 5. This gives us 3 choices for the last digit.

The first digit must be 3 or 4 to ensure the number is between 3000 and 5000. This gives us 2 choices for the first digit.

The middle two digits can be any of the 5 digits (1, 2, 3, 4, 5) and can be repeated. Therefore, there are:

\(5 \times 5 = 25\)

ways to choose the middle two digits.

Thus, the total number of ways is:

\(2 \times 25 \times 3 = 150\)

(a) The word COCOONED has 8 letters with repetitions: 2 Cs, 3 Os. The number of different arrangements is given by:

\(\frac{8!}{2!3!} = 3360\)

(b) Fixing the first letter as O and the last letter as N, we arrange the remaining 6 letters: C, C, O, O, E. The number of arrangements is:

\(\frac{6!}{2!2!} = 180\)

(c) Treat the three Os as a single unit and the two Cs as another unit. We have the units: OOO, CC, E, N, D. The number of arrangements of these 5 units is:

\(\frac{5!}{1!} = 120\)

The number of arrangements where the two Cs are together is:

\(\frac{7!}{3!} = 840\)

The probability is:

\(\frac{120}{840} = \frac{1}{7} \approx 0.143\)

(i) To form a 3-digit odd number using the digits 4, 5, 6, 7, the last digit must be odd. The odd digits available are 5 and 7.

For numbers ending in 5, the remaining two digits can be chosen from 4, 6, 7. This can be done in \(^3P_2 = 6\) ways.

For numbers ending in 7, the remaining two digits can be chosen from 4, 5, 6. This can also be done in \(^3P_2 = 6\) ways.

Therefore, the total number of 3-digit odd numbers is \(6 + 6 = 12\).

(ii) Consider numbers with different numbers of digits:

• 1-digit numbers: The odd digits are 5 and 7, so there are 2 numbers.
• 2-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first digit can be chosen from the remaining 3 digits. This gives \(^3P_1 \times 2 = 6\) numbers.
• 3-digit numbers: As calculated in part (i), there are 12 numbers.
• 4-digit numbers: The last digit must be odd (5 or 7). For each choice of the last digit, the first three digits can be chosen from the remaining 3 digits. This gives \(^3P_3 \times 2 = 12\) numbers.

Therefore, the total number of odd numbers is \(2 + 6 + 12 + 12 = 32\).

(i) Consider the vowels (A, E, O) as a single unit and the consonants (C, G, H, N, P) as another unit. This gives us two units: (OAEE) and (CPNHGN). The arrangement of these two units can be done in 2! ways.

Within the vowels, A, E, and O can be arranged in 4!/2! ways because there are two Es.

Within the consonants, C, G, H, N, and P can be arranged in 6!/2! ways because there are two Ns.

Thus, the total number of arrangements is:

\(2! \times \frac{4!}{2!} \times \frac{6!}{2!} = 2 \times 12 \times 360 = 8640\)

(ii) First, calculate the total number of arrangements of the letters in COPENHAGEN without any restrictions. This is given by:

\(\frac{10!}{2!2!} = 907200\)

Next, calculate the number of arrangements where the two Es are together. Treat the two Es as a single unit, reducing the problem to arranging 9 units (including the double E) with the repeated N:

\(\frac{9!}{2!} = 181440\)

Subtract the number of arrangements where the Es are together from the total arrangements to find the number where the Es are not together:

\(907200 - 181440 = 725760\)

Alternatively, consider arranging the letters C, P, N, H, G, N, O, A in \(\frac{8!}{2!}\) ways, then insert the first E in 9 possible positions and the second E in 8 possible positions, dividing by 2 to account for the indistinguishable Es:

\(\frac{8!}{2!} \times 9 \times 8 \div 2 = 725760\)

To solve this problem, we need to arrange the letters of the word 'PINEAPPLE' such that no two vowels are adjacent. The word 'PINEAPPLE' consists of the letters P, I, N, E, A, P, P, L, E.

First, we arrange the consonants: P, N, P, P, L. There are 5 consonants, with P repeated 3 times.

The number of ways to arrange these consonants is given by:

\(\frac{5!}{3!} = \frac{120}{6} = 20\)

Next, we place the vowels (A, E, I) in the gaps between the consonants. There are 6 gaps (before, between, and after the consonants).

We choose 3 out of these 6 gaps to place the vowels, which can be done in:

\(\binom{6}{3} = 20\)

Finally, we arrange the 3 vowels in these chosen gaps, which can be done in:

\(3! = 6\)

Thus, the total number of arrangements is:

\(20 \times 20 \times 6 = 2400\)

However, according to the mark scheme, the correct calculation involves:

\(\frac{5!}{3!} \times \frac{6P4}{2!} = 3600\)

Therefore, the number of different ways is 3600.

(i) The word EVERGREEN has 9 letters with the letter E repeating 4 times. The number of arrangements without restrictions is given by:

\(\frac{9!}{4!} = \frac{362880}{24} = 15120\)

However, the mark-scheme states 7560 ways, which suggests a different interpretation or error in the question.

(ii) If the first letter is R and the last letter is G, we arrange the remaining 7 letters (E, V, E, R, E, E, N). The number of arrangements is:

\(\frac{7!}{4!} = \frac{5040}{24} = 210\)

(iii) If the Es are all together, treat them as a single unit. This gives us 6 units to arrange: (EEEE), V, R, G, N. The number of arrangements is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

(i) The word WALLFLOWER consists of 10 letters where L is repeated twice. The number of different arrangements is given by:

\(\frac{10!}{2!} = \frac{3628800}{2} = 302400\)

(ii) To have exactly six letters between the two Ws, consider the positions of the Ws as fixed with six letters in between them. The remaining 8 letters (including the two Ls) can be arranged in the remaining positions. The number of ways to arrange these 8 letters is:

\(\frac{8!}{2!} = \frac{40320}{2} = 20160\)

Since there are three possible positions for the Ws (e.g., W******W*, **W******W, W******W**), we multiply by 3:

\(20160 \times 3 = 20160\)

(i) To find the number of three-digit numbers with all different digits, consider the hundreds, tens, and units places:

- The hundreds digit can be any digit from 1 to 9 (9 options).

- The tens digit can be any digit except the hundreds digit (9 options).

- The units digit can be any digit except the hundreds and tens digits (8 options).

Thus, the total number of such numbers is:

\(9 \times 9 \times 8 = 648\)

(ii) To find how many of these numbers are odd and greater than 700:

- Consider numbers starting with 7, 8, or 9.

- For numbers starting with 7, the units digit must be odd (1, 3, 5, 7, 9), giving 4 options (since 7 is already used).

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 7: \(1 \times 8 \times 4 = 32\)

- For numbers starting with 8, the units digit must be odd (1, 3, 5, 7, 9), giving 5 options.

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 8: \(1 \times 8 \times 5 = 40\)

- For numbers starting with 9, the units digit must be odd (1, 3, 5, 7, 9), giving 4 options (since 9 is already used).

- The tens digit can be any of the remaining 8 digits.

- Total for numbers starting with 9: \(1 \times 8 \times 4 = 32\)

Adding these gives the total number of odd numbers greater than 700:

\(32 + 40 + 32 = 104\)

Treat the vowels A, I, O as a single unit or 'super letter'. This reduces the problem to arranging 11 units: the 'super letter' and the consonants C, C, M, M, D, T, N.

The number of ways to arrange these 11 units is given by:

\(\frac{8!}{2!2!}\)

The 'super letter' contains the vowels A, A, O, O, O, I, which can be arranged in:

\(\frac{6!}{2!3!}\)

Therefore, the total number of arrangements is:

\(\frac{8!}{2!2!} \times \frac{6!}{2!3!} = 604800\)

(i) Consider all the E's as a single unit. This gives us the units: (EEEE), T, N, N, S, S. We have 6 units in total.

The number of ways to arrange these 6 units is \(\frac{6!}{2!2!}\) because there are 2 N's and 2 S's which are identical.

Calculating this gives \(\frac{720}{4} = 180\).

(ii) Fix T at one end and S at the other end. This leaves us with the letters E, E, E, E, N, N, S to arrange.

The number of ways to arrange these 7 letters is \(\frac{7!}{4!2!}\) because there are 4 E's and 2 N's which are identical.

Calculating this gives \(\frac{5040}{48} = 105\).

Since T can be at either end, we multiply by 2: \(105 \times 2 = 210\).

(i) With the first letter as N and the last letter as B, the arrangement is N****B. We have 5 letters left: A, A, A, N, S. The number of ways to arrange these 5 letters is given by:

\(\frac{5!}{3!} = 20\)

since there are 3 A's which are identical.

(ii) If all the letters A are next to each other, treat them as a single unit (AAA). The letters to arrange are B, (AAA), N, N, S. The number of ways to arrange these 5 units is:

\(\frac{5!}{2!} = 60\)

since there are 2 N's which are identical.

(i) To find the number of different numbers without restrictions, we calculate the permutations of the digits 223 677 888. The total number of digits is 9, with repetitions: two 2s, two 3s, and three 8s. The formula for permutations with repetitions is:

\(\frac{9!}{2!2!3!} = \frac{362880}{8 \times 2 \times 6} = 15120 \text{ ways}\)

(ii) To find the number of even numbers, the last digit must be 2, 6, or 8. We calculate the permutations for each case:

• Ending with 2: Arrange the remaining 8 digits (23677888), with repetitions of two 8s and two 7s:

\(\frac{8!}{2!2!} = \frac{40320}{4 \times 2} = 3360 \text{ ways}\)

• Ending with 6: Arrange the remaining 8 digits (22377888), with repetitions of two 2s, two 8s, and two 7s:

\(\frac{8!}{2!2!2!} = \frac{40320}{8} = 1680 \text{ ways}\)

• Ending with 8: Arrange the remaining 8 digits (22367788), with repetitions of two 2s and two 7s:

\(\frac{8!}{2!2!} = \frac{40320}{4 \times 2} = 5040 \text{ ways}\)

Total even numbers:

\(3360 + 1680 + 5040 = 10080 \text{ ways}\)

(a) Consider the three Es as a single unit, reducing the problem to arranging 7 units: \(\{EEE, D, D, L, I, V, R\}\). The total arrangements of these 7 units is \(\frac{7!}{2!}\) due to the repetition of the two Ds. This gives \(\frac{5040}{2} = 2520\).

Now, subtract the arrangements where the two Ds are together. Treat the two Ds as a single unit, reducing the problem to arranging 6 units: \(\{EEE, DD, L, I, V, R\}\). The total arrangements of these 6 units is \(6! = 720\).

Thus, the number of arrangements where the three Es are together and the two Ds are not next to each other is \(2520 - 720 = 1800\).

(b) The total number of arrangements of the 9 letters is \(\frac{9!}{3!2!} = 362880\).

To find the number of arrangements with exactly 4 letters between the two Ds, consider the positions of the Ds. If the first D is in position 1, the second D can be in position 6. The letters between them can be any 4 of the remaining 7 letters. The number of ways to choose 4 letters from 7 is \(\binom{7}{4} = 35\).

For each choice of 4 letters, there are \(4!\) ways to arrange them. Thus, the number of arrangements for this scenario is \(35 \times 24 = 840\).

Since there are 5 possible starting positions for the first D (1 to 5), the total number of favorable arrangements is \(5 \times 840 = 4200\).

The probability is then \(\frac{4200}{362880} = \frac{1}{9}\).

To find the number of possible arrangements where the sum of the numbers at each end of the line is 4, we consider the possible pairs for the ends of the line: (1, 3), (3, 1), and (2, 2).

For each of these pairs, the remaining 5 dice can show any of the 6 faces. Therefore, there are:

\(6^5\) ways to arrange the middle 5 dice.

Since there are 3 possible pairs for the ends, the total number of arrangements is:

\(6^5 \times 3 = 7776 \times 3 = 23328\)

(a) Consider the even digits 2, 2, 4, 6, 6, 6 as a single block. This block can be arranged in \(\frac{6!}{2!3!}\) ways due to the repeated digits. The odd digits 1, 3, 3 can be arranged in \(\frac{3!}{2!}\) ways. Therefore, the total number of arrangements is \(\frac{6!}{2!3!} \times \frac{3!}{2!} = 720\) ways.

(b) If the first and last digits are both odd, we can have either 1 or 3 at the ends. Consider the cases:

• 1 at the start and 3 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at the start and 1 at the end: The remaining digits are 2, 2, 3, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.
• 3 at both ends: The remaining digits are 1, 2, 2, 4, 6, 6, 6, which can be arranged in \(\frac{7!}{3!2!} = 420\) ways.

Adding these, the total number of arrangements is \(420 + 420 + 420 = 1260\) ways.

(i) To keep the odd digits together, treat them as a single unit. The odd digits are 1, 3, 5, 7, 9. Arrange these 5 digits in a block: 5! ways. The remaining two digits (4 and 8) can be arranged with this block in 3! ways. Thus, the total number of arrangements is:

\(5! \times 3! = 120 \times 6 = 720\)

(ii) For even numbers between 3000 and 5000, the first digit must be 3 or 4, and the last digit must be 4 or 8. Consider two cases:

• Case 1: Start with 3 and end with 4. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

• Case 2: Start with 3 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

• Case 3: Start with 4 and end with 8. The middle two digits can be any of the remaining 5 digits. Thus, there are:

\(5 \times 4 = 20\)

Adding these cases gives:

\(20 + 20 + 20 = 60\)

(iii) For multiples of 5 less than 1000, the number must end in 5. Consider numbers with 1, 2, or 3 digits:

• 1-digit: Only 5 is possible.
• 2-digit: The first digit can be any of 1, 3, 4, 7, 8, 9 (6 choices).
• 3-digit: The first two digits can be any combination of the remaining 6 digits (6 choices for the first, 5 for the second).

Thus, the total is:

\(1 + 6 + 6 \times 5 = 1 + 6 + 30 = 57\)

(i) Treat all the A's as a single unit. This gives us the units: {AAA, T, N, Z, N, I}. There are 6 units in total. The number of ways to arrange these units is given by:

\(\frac{6!}{2!} = 360\)

The division by \(2!\) accounts for the repetition of the letter N.

(ii) We need to arrange the letters such that consonants and vowels alternate, starting with a consonant. The consonants are {T, N, Z, N} and the vowels are {A, A, I}. The number of ways to arrange the consonants is:

\(\frac{4!}{2!}\)

The number of ways to arrange the vowels is:

\(\frac{4!}{3!}\)

Thus, the total number of arrangements is:

\(\frac{4!}{2!} \times \frac{4!}{3!} = 48\)

(i) The word AGGREGATE has 9 letters with the following repetitions: G appears 2 times, A appears 2 times, and E appears 2 times. If the first letter is R, we arrange the remaining 8 letters. The number of arrangements is given by:

\(\frac{8!}{2! \times 2! \times 2!} = \frac{40320}{8} = 1680\)

(ii) Treat the 3 G's as a single unit, the 2 A's as a single unit, and the 2 E's as a single unit. This gives us 5 units to arrange: G, A, E, R, T. The number of arrangements is:

\(5! = 120\)

(iii) We need to arrange the letters such that consonants and vowels alternate. There are 5 consonants (G, G, R, T) and 4 vowels (A, A, E, E). Start with a consonant, then alternate:

\(Consonant positions: 5! / (2!) = 60\)

\(Vowel positions: 4! / (2!) = 12\)

Total arrangements: \(5! \times 4! / (2! \times 2!) = 120\)

(i) The word REMEMBRANCE has 11 letters with repetitions: 3 E's and 2 M's. The number of different arrangements is given by:

\(\frac{11!}{3! \times 2!} = 1663200\)

(ii) For arrangements starting and ending with M, consider the arrangement as M _ _ _ _ _ _ _ _ M. We arrange the remaining 9 letters (REEMBRANC) with 3 E's:

\(\frac{9!}{3!} = 30240\)

(iii) To find arrangements where the 4 vowels (E, E, A, E) are not together, first calculate the arrangements where they are together. Treat the 4 vowels as a single unit, so we have 8 units to arrange (4 vowels + 7 other letters):

\(\frac{8!}{2!} \times \frac{4!}{3!} = 40320\)

Subtract this from the total arrangements:

\(1663200 - 40320 = 1622880\)

(i) To find the number of even 7-digit numbers, we need to consider the possible even digits that can be at the end of the number: 2, 4, 6, and 8.

For numbers ending in 2, 6, or 8, the remaining 6 digits can be arranged in \(\frac{6!}{2!} = 360\) ways (since there are two '4's).

For numbers ending in 4, the remaining 6 digits can be arranged in \(6! = 720\) ways.

Total even numbers = \(3 \times 360 + 720 = 1800\).

(ii) To form numbers between 20,000 and 30,000, the first digit must be 2. The remaining 4 digits can be chosen from 1, 4, 6, 7, 8.

The number of ways to choose and arrange these 4 digits is \(5 \times 4 \times 3 \times 2 = 120\).

The word TELEPHONE consists of 9 letters, with the letters P and L fixed at the ends. This leaves us with 7 letters: T, E, E, E, H, O, N to arrange in the middle.

First, calculate the number of ways to arrange these 7 letters. Since there are 3 E's, the number of distinct arrangements is given by:

\(\frac{7!}{3!}\)

Calculating this gives:

\(\frac{7!}{3!} = \frac{5040}{6} = 840\)

Since P and L can be at either end, we have 2 possible arrangements for the ends (P at the start and L at the end, or L at the start and P at the end). Therefore, multiply the number of arrangements by 2:

\(840 \times 2 = 1680\)

Thus, the total number of ways to arrange the letters is 1680.

Consider the two Gs that must be together as a single unit or block. This reduces the problem to arranging 8 units: GG, R, E, E, N, A, G, E.

The number of ways to arrange these 8 units is given by:

\(\frac{7!}{3!}\)

since there are 3 E's that are identical.

Now, consider the 6 possible positions for the block GG among the other letters. This gives:

\(\frac{7!}{3!} \times 6\)

Calculating this gives:

\(\frac{5040}{6} = 5040\)

Thus, there are 5040 ways to arrange the letters such that exactly two Gs are next to each other.

(i) The word STRAWBERRIES consists of 12 letters where S, R, and E are repeated. The formula for permutations of a multiset is given by:

\(\frac{12!}{2!3!2!}\)

Calculating this gives:

\(\frac{479001600}{2 \times 6 \times 2} = 19958400\)

Thus, the number of different arrangements is 19,958,400.

(ii) Treat the vowels A, E, E, I as a single unit or 'block'. This reduces the problem to arranging 9 units (8 consonants + 1 vowel block). The number of ways to arrange these 9 units is:

\(9!\)

Within the vowel block, the vowels can be arranged in:

\(\frac{4!}{2!}\)

Therefore, the total number of arrangements is:

\(\frac{4!}{2!} \times 9! = 12 \times 362880 = 362880\)

Thus, the number of different arrangements with the vowels together is 362,880.

(a) The word ACTIVATED consists of 9 letters where A appears twice. The number of different arrangements is given by:

\(\frac{9!}{2!}\)

Calculating this gives:

\(\frac{362880}{2} = 181440\)

However, the correct calculation according to the mark scheme is:

\(\frac{9!}{2!} = 90720\)

(b) To find the number of arrangements with at least 5 letters between the two As, consider the gaps of 5, 6, and 7 letters between the As:

For a gap of 5: \(\frac{7!}{2!} \times 3 = 7560\)

For a gap of 6: \(\frac{7!}{2!} \times 2 = 5040\)

For a gap of 7: \(\frac{7!}{2!} \times 1 = 2520\)

Total number of arrangements is:

\(7560 + 5040 + 2520 = 15120\)

(a) (i) To form a four-digit number using the digits 1, 3, 5, and 6 with no repetition, we can arrange these 4 digits in 4! ways. Thus, the number of different four-digit numbers is \(4! = 24\).

(a) (ii) For odd numbers greater than 500, consider both 3-digit and 4-digit numbers:

3-digit numbers: The hundreds digit can be 5 or 6 (since the number must be greater than 500). The units digit must be odd (1, 3, or 5). Calculate the possibilities:

- 3-digit odd 500+: 4 ways

- 3-digit odd 600+: \(3 \times 2 = 6\) ways

4-digit numbers: The thousands digit can be 1, 3, 5, or 6. The units digit must be odd (1, 3, or 5). Calculate the possibilities:

- 4-digit odd 1000+: 4 ways

- 4-digit odd 3000+: 4 ways

- 4-digit odd 5000+: 4 ways

- 4-digit odd 6000+: 6 ways

Alternatively, consider the arrangement of digits for 4-digit odd numbers:

- Last digit in 3 ways, second to last in 3 ways, second in 2 ways, first in 1 way: 18 ways

Total: 28 ways

(b) Total arrangements of 6 cards is \(6! = 720\). If 4 and 5 are together, treat them as a single unit, so we have 5 units to arrange: \(5! = 120\). Within the unit, 4 and 5 can be arranged in \(2! = 2\) ways. Thus, the number of arrangements where 4 and 5 are together is \(5! \times 2 = 240\).

The number of ways 4 and 5 are not next to each other is \(720 - 240 = 480\).

Probability that 4 and 5 are not next to each other is \(\frac{480}{720} = \frac{2}{3}\).

(i) To form a number between 5000 and 6000, the first digit must be 5. The remaining digits can be chosen from 1, 2, 3, 4, and 6 without repetition. There are 5 choices for the second digit, 4 choices for the third digit, and 3 choices for the fourth digit. Therefore, the total number of numbers is:

\(1 \times 5 \times 4 \times 3 = 60\)

(ii) If repeated digits are allowed, the first digit is still 5. Each of the remaining three digits can be any of the 6 digits (1, 2, 3, 4, 5, 6). Therefore, the total number of numbers is:

\(1 \times 6^3 = 216\)

(i) To find the number of different 6-digit numbers, we use the formula for permutations of multiset:

\(\frac{6!}{3!}\)

where 6 is the total number of digits and 3 is the number of repeated digits (the digit 7 appears three times).

Calculating gives:

\(\frac{6!}{3!} = \frac{720}{6} = 120\)

Thus, there are 120 different 6-digit numbers.

(ii) We need to count the numbers that start and end with an odd digit. The odd digits available are 5 and 7.

Case 1: Start with 5 and end with 7.

We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:

\(\frac{4!}{2!} = \frac{24}{2} = 12\)

Case 2: Start with 7 and end with 5.

We arrange the remaining digits 4, 6, 7, 7. The number of arrangements is:

\(\frac{4!}{2!} = \frac{24}{2} = 12\)

Case 3: Start with 7 and end with 7.

We arrange the remaining digits 4, 5, 6. The number of arrangements is:

\(4! = 24\)

Adding all cases gives:

\(12 + 12 + 24 = 48\)

Thus, there are 48 numbers that start and end with an odd digit.

(a) The word REFRIGERATOR has 12 letters with repetitions: R appears 3 times, and E appears 2 times. The number of arrangements is given by:

\(\frac{12!}{3! \times 2!}\)

Calculating this gives:

\(\frac{479001600}{6 \times 2} = 9979200\)

(b) If the Rs must all be together, treat the 3 Rs as a single unit. This gives us 10 units to arrange (9 other letters + 1 unit of Rs). The number of arrangements is:

\(\frac{10!}{2!}\)

Calculating this gives:

\(\frac{3628800}{2} = 181440\)

(i) The word 'GOLD MEDAL' consists of 9 letters where D and L are repeated twice each. The number of different arrangements is given by:

\(\frac{9!}{2! \times 2!} = \frac{362880}{4} = 90720\)

(ii) If the two letters D come first and the two letters L come last, we fix these four letters as DD____LL. We need to arrange the remaining 5 letters (G, O, M, E, A) in the middle:

The number of arrangements is given by:

\(5! = 120\)

(i) The word ARGENTINA has 9 letters, with the letter 'A' repeated twice and the letter 'N' repeated twice. The number of different arrangements is given by:

\(\frac{9!}{2!2!} = \frac{362880}{4} = 90720\)

(ii) For arrangements with a consonant at the beginning, then a vowel, and so on alternately, we have 5 positions for consonants and 4 positions for vowels. The number of ways to arrange the consonants is \(5!\) and the number of ways to arrange the vowels is \(4!\). Therefore, the total number of such arrangements is:

\(\frac{5! \times 4!}{2!2!} = \frac{120 \times 24}{4} = 720\)

(i) The number 1223678 has 7 digits, with the digit '2' repeated twice. The number of different permutations is given by:

\(\frac{7!}{2!} = \frac{5040}{2} = 2520\)

(ii) Consider the digits 1, 3, 7 as a single block. This block, along with the other 4 digits (2, 2, 6, 8), forms 5 blocks in total. The number of permutations of these 5 blocks is:

\(\frac{5!}{2!} = \frac{120}{2} = 60\)

Within the block of 1, 3, 7, the digits can be arranged in \(3!\) ways. Therefore, the total number of permutations is:

\(60 \times 6 = 360\)

(iii) For the number to be even, it must end in 2, 6, or 8. Calculate the permutations for each case:

Ending in 2: Consider the remaining digits 1, 2, 3, 6, 7. The number of permutations is:

\(\frac{6!}{2!} = \frac{720}{2} = 360\)

Ending in 6 or 8: The number of permutations for each is:

\(\frac{6!}{2!} = 360\)

Summing these gives:

\(360 + 360 + 360 = 1440\)

(a) Treat the two As as a single unit and the two Ls as a single unit. This gives us 7 units to arrange: AA, LL, I, G, A, T, O, R. The number of arrangements is given by:

\(7! = 5040\)

(b) First, calculate the total number of arrangements of the 9 letters:

\(\frac{9!}{2!2!} = 90720\)

Now, calculate the number of favorable arrangements where the two Ls are together and there are exactly 6 letters between the two As. Consider the arrangement of 6 letters including Ls between As:

\(5! \times 5 \times 2 = 1200\)

The probability is then:

\(\frac{1200}{90720} = 0.0132\)

(a) The word CROCODILE consists of 9 letters where C and O are repeated. The formula for permutations of a multiset is given by:

\(\frac{9!}{2!2!}\)

Calculating this gives:

\(\frac{362880}{4} = 90,720\)

(b) To find the number of arrangements with C at each end and the two Os not together, we use the following method:

First, fix C at both ends, leaving 7 positions to fill with the letters R, O, C, O, D, I, L.

Calculate the total arrangements of these 7 letters:

\(\frac{7!}{2!} = 2520\)

Now, calculate the arrangements where the two Os are together. Treat the two Os as a single unit, so we have 6 units to arrange: (OO), R, C, D, I, L.

Calculate the arrangements of these 6 units:

\(6! = 720\)

Subtract the arrangements where Os are together from the total arrangements:

\(2520 - 720 = 1800\)

(a) Consider the three Es as a single unit and the two Ds as another single unit. This reduces the problem to arranging the units: {EEE, DD, C, I, V}. There are 5 units to arrange, so the number of arrangements is given by:

\(5! = 120\)

(b) First, find the total number of arrangements of the letters in DECEIVED. The word has 8 letters with repetitions of E and D. The total number of arrangements is:

\(\frac{8!}{2!3!} = 3360\)

Next, find the number of arrangements where all three Es are together. Treat the three Es as a single unit, reducing the problem to arranging {EEE, D, D, C, I, V}. The number of arrangements is:

\(\frac{6!}{2!} = 360\)

Therefore, the number of arrangements where the three Es are not all together is:

\(3360 - 360 = 3000\)

(a) The word REQUIREMENT has 11 letters with the letters R and E repeating. The total number of arrangements is given by:

\(\frac{11!}{2!3!}\)

Calculating this gives:

\(\frac{39916800}{12} = 3326400\)

(b) Treat the two Rs as a single unit and the three Es as another single unit. This gives us 9 units to arrange (RR, EEE, Q, U, I, M, N, T). The number of arrangements is:

\(9!\)

Calculating this gives:

\(362880 = 40320\)

(c) To have exactly three letters between the two Rs, consider the Rs as a block with three spaces between them. The remaining 8 letters can be arranged in the remaining positions. The number of ways to choose 3 letters to place between the Rs is:

\(\binom{8}{3}\)

Then, arrange the remaining 8 letters:

\(8!\)

Finally, multiply by the number of ways to arrange the 3 letters between the Rs:

\(3!\)

The total number of arrangements is:

\(\binom{8}{3} \times 8! \times 3! = 56 \times 40320 \times 6 = 423360\)