Exercise 6.3 answers: Permutations and Combinations

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1. There are 9 digits to choose from (1 to 9) and 3 places to fill, with no repeats.
2. Hundreds place: 9 choices. Tens place: 8 digits left. Units place: 7 digits left.
3. Multiply: 9 × 8 × 7 = 504.

Answer504

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1. A 4-digit number cannot start with 0, so the first place has 9 choices (1 to 9).
2. Once the first digit is fixed, 9 digits remain for the second place (the 8 unused digits plus 0).
3. Third place: 8 digits left. Fourth place: 7 digits left.
4. Multiply: 9 × 9 × 8 × 7 = 4536.

Answer4536

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1. A number is even when its units digit is even, so fill the units place first.
2. Units place: only 2, 4 or 6 can go here — 3 choices.
3. Hundreds place: 5 digits are left. Tens place: 4 digits are left.
4. Multiply: 3 × 5 × 4 = 60.

Answer60

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1. There are 5 digits and 4 places to fill, with no digit repeated.
2. Thousands place: 5 choices. Hundreds: 4 digits left. Tens: 3 digits left. Units: 2 digits left.
3. Multiply: 5 × 4 × 3 × 2 = 120 numbers in all.
4. For an even number, the units digit must be 2 or 4 — 2 choices.
5. Fill the units place first: 2 choices there, then 4 digits left for the thousands place, 3 for the hundreds place, 2 for the tens place.
6. Multiply: 2 × 4 × 3 × 2 = 48 of these numbers are even.

Answer120 four-digit numbers in all, of which 48 are even.

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1. The chairman and vice chairman are two different posts, and nobody can hold both.
2. Chairman: 8 choices. Vice chairman: 7 people are left to choose from.
3. Multiply: 8 × 7 = 56.

Answer56

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1. n−1P3 = (n−1)!/(n−4)! and nP4 = n!/(n−4)!.
2. Divide one by the other: the (n−4)! cancels, leaving (n−1)!/n!, which is 1/n.
3. Set 1/n = 1/9, so n = 9.

Answer9

Watch this explained “Solving for n”, 9:35 into The closed formula, and how allowing repeats changes the count entirely

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(i) 5Pr = 2 6Pr–1

1. Write both sides as factorial quotients: 5Pr = 120/(5−r)!, and 6P(r−1) = 720/(7−r)!.
2. The equation becomes 120/(5−r)! = 2 × 720/(7−r)!, so (7−r)!/(5−r)! = 12.
3. (7−r)!/(5−r)! = (7−r)(6−r), so (7−r)(6−r) = 12, giving r² − 13r + 30 = 0.
4. Solving gives r = 3 or r = 10; r = 10 is impossible since r cannot exceed 5, so r = 3.

Answer3

(ii) 5Pr = 6Pr–1

1. Write both sides as factorial quotients: 5Pr = 120/(5−r)!, and 6P(r−1) = 720/(7−r)!.
2. The equation becomes 120/(5−r)! = 720/(7−r)!, so (7−r)(6−r) = 6, giving r² − 13r + 36 = 0.
3. Solving gives r = 4 or r = 9; r = 9 is impossible since r cannot exceed 5, so r = 4.

Answer4

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1. EQUATION has 8 letters, and all 8 letters are different.
2. Using all 8 letters exactly once means arranging all 8 of them in a row.
3. The number of arrangements of 8 different letters is 8! = 40320.

Answer40320

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(i) 4 letters are used at a time

1. MONDAY has 6 different letters.
2. Filling 4 places from 6 letters, with no repeats: 6 × 5 × 4 × 3.
3. That comes to 360.

Answer360

(ii) all letters are used at a time

1. Using all 6 different letters means arranging all of them in a row.
2. The number of arrangements of 6 different letters is 6! = 720.

Answer720

(iii) all letters are used but first letter is a vowel

1. MONDAY has 2 vowels, O and A.
2. The first letter must be a vowel: 2 choices.
3. The remaining 5 letters fill the other 5 places in any order: 5! = 120 ways.
4. Multiply: 2 × 120 = 240.

Answer240

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1. MISSISSIPPI has 11 letters: M once, I four times, S four times, P twice.
2. All distinct arrangements: 11!/(4! × 4! × 2!) = 34650.
3. To count arrangements with all four I's together, treat the block of four I's as a single object.
4. That leaves 8 objects (the I-block, M, 4 S's, 2 P's) to arrange: 8!/(4! × 2!) = 840.
5. Arrangements where the four I's do NOT all come together: 34650 − 840 = 33810.

Answer33810

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(i) words start with P and end with S

1. PERMUTATIONS has 12 letters, and the letter T is the only one that repeats — it appears twice.
2. Fix P at the first place and S at the last place.
3. That leaves 10 letters in between: E, R, M, U, T, A, T, I, O, N — with T still repeated twice.
4. Arrange these 10 letters and divide by 2! because the two T's cannot be told apart: 10!/2!.
5. 10!/2! = 3628800/2 = 1814400.

Answer1814400

(ii) vowels are all together

1. The vowels in PERMUTATIONS are E, U, A, I, O — 5 different vowels.
2. Tie all 5 vowels into one block. Now there are 7 consonants (P, R, M, T, T, N, S) plus this 1 block — 8 items to arrange, with T still repeated twice.
3. Arrange these 8 items: 8!/2! = 20160.
4. Inside the block, the 5 different vowels can be arranged in 5! = 120 ways.
5. Multiply: 20160 × 120 = 2419200.

Answer2419200

(iii) there are always 4 letters between P and S

1. Number the 12 positions 1 to 12.
2. "4 letters between P and S" means their positions are 5 apart, like position 1 and position 6.
3. Pairs of positions that are 5 apart: (1,6), (2,7), (3,8), (4,9), (5,10), (6,11), (7,12) — 7 pairs, and P or S can sit in the earlier spot, so 14 ways to place P and S.
4. The other 10 letters (with T repeated twice) fill the remaining 10 places in 10!/2! = 1814400 ways.
5. Multiply: 14 × 1814400 = 25401600.

Answer25401600

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