S4 Mathematics Practice Book · Unit 13: Combinatorics

Unit 13 Combinatorics

Key unit competence: Use arrangements and combinations in determining the number of ways a random experiment occurs.

Key ideas

Counting tools: lists, Venn diagrams, tree diagrams, contingency tables.
Multiplication principle: if one task can be done in $m$ ways and another in $n$ ways, both together can be done in $m \times n$ ways. Addition principle: either one or the other (not both): $m + n$ ways.
Arrangements with repetition: $n^{p}$ ways to fill $p$ places from $n$ objects.
Permutations (order matters, no repetition): $P (n, r) =^{n} P_{r} = \frac{n !}{( n - r )!}$ ; all $n$ objects: $n!$ . With identical items: $\frac{n !}{a ! b ! \dots}$ .
Combinations (order does not matter): $C (n, r) = (r n) = \frac{n !}{r ! ( n - r )!}$ . Properties: $(r n) = (n - r n)$ , $(r n) + (r + 1 n) = (r + 1 n + 1)$ (Pascal's triangle).
Binomial theorem: $(a + b)^{n} = \sum_{k = 0}^{n} (k n) a^{n - k} b^{k}$ . The general term is $T_{k + 1} = (k n) a^{n - k} b^{k}$ .

Formulas to know

n! = n (n - 1) \dots 2 \cdot 1, 0! = 1

P (n, r) = \frac{n !}{( n - r )!}

(r n) = \frac{n !}{r ! ( n - r )!}

(a + b)^{n} = k = 0 \sum n (k n) a^{n - k} b^{k}

Common misconceptions

"Choosing a committee is a permutation." No: the order of members does not matter, so use combinations.
" $0! = 0$ ." No: $0! = 1$ .
"The number of 4-digit PINs from 10 digits is $P (10, 4)$ ." Only if digits cannot repeat; with repetition it is $1 0^{4}$ .

Worked examples

Example 1. How many ways can a chairperson, secretary and treasurer be chosen from 8 students?

Solution

Order matters: $P (8, 3) = 8 \times 7 \times 6 = 336$ .

Example 2. Expand $(x + 2)^{4}$ .

Solution

Coefficients 1, 4, 6, 4, 1: $x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16$ .

Practice questions

Easier and harder questions are mixed, just like in a real exam.

How many ways can the first three places be awarded in a race of 10 runners?
Show answer
$P (10, 3) = 720$ .
Common mistake: Using $(3 10)$ (positions are ordered).
Car plates have 3 letters (A–Z) followed by 3 digits. How many plates are possible?
Show answer
$2 6^{3} \times 1 0^{3} = 17576000$ .
Common mistake: Assuming no repetition.
Expand $(x + 2)^{4}$ using Pascal's triangle.
Show answer
$x^{4} + 8 x^{3} + 24 x^{2} + 32 x + 16$ .
Common mistake: Forgetting to raise 2 to the right powers.
How many ways can 4 people sit around a round table?
Show answer
$(4 - 1)! = 6$ (rotations are the same).
Common mistake: Answering $4! = 24$ .
How many 4-letter "words" can be formed from the letters of RWANDA without repetition (treat the two As as identical)?
Show answer
Count by cases. Letters R, W, A, N, D, A. With at most one A: 5 distinct letters choose 4 and arrange: $P (5, 4) = 120$ . With both As: choose 2 of R, W, N, D ( $(2 4) = 6$ ) and arrange with A, A: $6 \times \frac{4 !}{2 !} = 72$ . Total 192.
Common mistake: Using $P (6, 4) = 360$ (this counts the As as different).
Find the 4th term of $(2 x - 3)^{6}$ .
Show answer
$T_{4} = (3 6) (2 x)^{3} (- 3)^{3} = 20 \times 8 x^{3} \times (- 27) = - 4320 x^{3}$ .
Common mistake: Using $(4 6)$ for the 4th term.
A team of 5 is chosen from 7 boys and 4 girls with at least 2 girls. How many ways?
Show answer
2 girls: $(2 4) (3 7) = 210$ ; 3 girls: $(3 4) (2 7) = 84$ ; 4 girls: $(4 4) (1 7) = 7$ . Total 301.
Common mistake: Using $(2 4) (3 9)$ (overcounting).
Prove that $(r n) = (n - r n)$ .
Show answer
$(n - r n) = \frac{n !}{( n - r )! r !} = (r n)$ . Choosing $r$ to take is the same as choosing $n - r$ to leave.
Common mistake: Testing only one value of $n$ .

Matching and drag-and-drop

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. How many even 3-digit numbers can be made from ${1, 2, 3, 4, 5}$ without repetition?	a. $24$
2. Find the term in $x^{4}$ in the expansion of $(x - 2)^{6}$ .	b. $n = 8$
3. Find $(3 10) + (4 10)$ .	c. $60 x^{4}$
4. Solve $P (n, 2) = 56$ .	d. $330$

Show answer

1 → a ( $24$ ) · 2 → c ( $60 x^{4}$ ) · 3 → d ( $330$ ) · 4 → b ( $n = 8$ )

Step 1. Last digit even: 2 or 4 (2 ways).
Step 2. Then 4 choices for the first digit and 3 for the middle: $2 \times 4 \times 3 = 24$ .
Step 1. General term $(k 6) x^{6 - k} (- 2)^{k}$ ; $x^{4}$ needs $k = 2$ .
Step 2. $(2 6) \times 4 = 15 \times 4 = 60$ .
Step 1. Pascal: $(r n) + (r + 1 n) = (r + 1 n + 1)$ .
Step 2. $(4 11) = 330$ (check: $120 + 210 = 330$ ).
Step 1. $P (n, 2) = n (n - 1) = 56$ .
Step 2. $n^{2} - n - 56 = 0$ , $(n - 8) (n + 7) = 0$ , so $n = 8$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$35$ · $110$ · $45$ · $34650$
1. In how many ways can a captain and a vice-captain be chosen from 11 players? →
2. How many diagonals does a polygon with 10 sides have? →
3. How many arrangements of MISSISSIPPI are there? →
Show answer
1: $110$ · 2: $35$ · 3: $34650$
- Step 1. The roles are different, so order matters.
  Step 2. $P (11, 2) = 11 \times 10 = 110$ .
- Step 1. Segments joining two vertices: $(2 10) = 45$ .
  Step 2. Remove the 10 sides: $45 - 10 = 35$ .
- Step 1. 11 letters: I four times, S four times, P twice.
  Step 2. $\frac{11 !}{4 ! 4 ! 2 !} = 34650$ .

Matching

Match each item in column A with the correct answer in column B. Write the letter next to each number.

Column A	Column B
1. In how many ways can 3 students be chosen from 8?	a. $56$
2. Find the middle term of $(x + \frac{1}{x})^{4}$ .	b. $150$
3. How many 4-letter arrangements can be made from the letters of MATHS (no repetition)?	c. $6$
4. A committee of 4 is chosen from 6 men and 5 women. How many have exactly 2 women?	d. $120$

Show answer

1 → a ( $56$ ) · 2 → c ( $6$ ) · 3 → d ( $120$ ) · 4 → b ( $150$ )

Step 1. Order does not matter: $(3 8)$ .
Step 2. $\frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$ .
Step 1. 5 terms: the middle one is $k = 2$ .
Step 2. $(2 4) x^{2} (\frac{1}{x})^{2} = 6$ .
Step 1. 5 choices, then 4, then 3, then 2.
Step 2. $5 \times 4 \times 3 \times 2 = 120$ .
Step 1. Choose 2 women: $(2 5) = 10$ ; 2 men: $(2 6) = 15$ .
Step 2. $10 \times 15 = 150$ .

Drag and drop
Choose the correct word or phrase from the box for each gap. One word in the box is not needed.
$240$ · $40$ · $48$ · $74$
1. A committee of 3 is chosen from 5 boys and 4 girls. How many include at least one girl? →
2. Find the coefficient of $x^{2}$ in $(1 + 2 x)^{5}$ . →
3. In how many ways can 6 books be arranged on a shelf if 2 particular books must be together? →
Show answer
1: $74$ · 2: $40$ · 3: $240$
- Step 1. All committees: $(3 9) = 84$ .
  Step 2. With no girl: $(3 5) = 10$ .
  Step 3. $84 - 10 = 74$ .
- Step 1. Term $(2 5) (2 x)^{2}$ .
  Step 2. $10 \times 4 x^{2} = 40 x^{2}$ .
- Step 1. Glue the two books into one block: 5 items, $5! = 120$ .
  Step 2. The two books can swap: $\times 2 = 240$ .

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