Unit 5 Review Exercises

Counting Principles

All questions are AHL. Paper 1 means no calculator; Paper 2 means a GDC is allowed.

1 Paper 1 In how many ways can the letters of SQUARE be arranged

(a) with no restriction,

(b) if the word must start and end with a vowel?

Solution

(a) \(6! = 720\).

(b) Vowels U, A, E: ends \(3 \times 2 = 6\); middle \(4! = 24\). Total \(144\). See 5.2.

2 Paper 1 How many 4-digit PIN codes have four different digits (0–9 allowed in any place)?

Solution

\({}^{10}\mathrm{P}_4 = 10 \times 9 \times 8 \times 7 = 5040\). See 5.1.

3 Paper 1 How many 5-digit numbers with different digits, made from the digits 1–9, are divisible by 5?

Solution

Last digit must be 5. The other four places: \({}^8\mathrm{P}_4 = 1680\). See 5.2.

4 Paper 2 A committee of 5 is chosen from 8 teachers and 6 students. How many committees contain at least 2 students?

Solution

Total \(-\) (0 students) \(-\) (1 student):

\[ \binom{14}{5} - \binom{8}{5} - \binom{8}{4}\binom{6}{1} = 2002 - 56 - 420 = 1526 . \]

See 5.3.

5 Paper 1 Three boys and three girls stand in a row so that boys and girls alternate. In how many ways can they stand?

Solution

Pattern BGBGBG or GBGBGB: 2 ways. Boys \(3!\), girls \(3!\). Total \(2 \times 6 \times 6 = 72\). See 5.2.

6 Paper 1 Solve \(\dbinom{n}{3} = 2\dbinom{n}{2}\), where \(n \ge 3\).

Solution

\(\dfrac{n(n - 1)(n - 2)}{6} = n(n - 1) \Rightarrow \dfrac{n - 2}{6} = 1 \Rightarrow n = 8\). See 5.3.

7 Paper 1 Two maths books are chosen from 6, and three science books from 5. The five books are then arranged on a shelf. In how many ways can this be done?

Solution

\(\dbinom{6}{2}\dbinom{5}{3} \times 5! = 15 \times 10 \times 120 = 18\,000\). See 5.3.

8 Paper 1 Four students are chosen at random from 5 boys and 7 girls. Find the probability that 2 boys and 2 girls are chosen.

Solution

\(\dfrac{\binom{5}{2}\binom{7}{2}}{\binom{12}{4}} = \dfrac{10 \times 21}{495} = \dfrac{14}{33}\). See 5.3.

9 Paper 1 Extension How many different arrangements are there of the letters of LETTER?

Solution

E×2, T×2: \(\dfrac{6!}{2!\,2!} = 180\). See 5.2.

10 Paper 1 Extension Six people sit at a round table. In how many ways can they sit

(a) with no restriction,

(b) if two particular people sit together?

Solution

(a) \(5! = 120\). \(\quad\) (b) \(4! \times 2 = 48\). See 5.2.