Past Exam Questions

(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:

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