Past Exam Questions

(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.