Mathematics 9709 · Probability & Statistics 1

Permutations vs combinations

This is the main decision in the chapter: does order matter?
If order matters, use $P$ . If order does not matter, use $C$ .
A combination is a permutation with the extra orderings divided out.

P arranges; C chooses

Start with one question: would swapping the chosen objects make a new answer?
If swapping changes the answer, order matters, so use $P$ . If swapping does not change the answer, order is ignored, so use $C$ .
Case splits are a later step. First learn the small difference between $AB$ and $BA$ .

${}^nC_r=\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!(n-r)!}$ . The $r!$ removes the order of the chosen group.

Question wording	Order?	Use
arrange, code, row, queue, seats, podium	yes	${}^nP_r$
choose, select, team, committee, hand, group	no	${}^nC_r$
several independent choices	depends on each choice	multiply the counts
at least / at most	case split	add legal cases only

choose: ⁿCr = 10×r! = 6=arrange: ⁿPr = 60

Worked example

First rung: three letters

Arrange 2 letters from A, B and C. AB and BA are different, so order matters: ${}^3P_2=3\times2=6$ .
Choose 2 letters from A, B and C. AB and BA are the same pair, so order does not matter: ${}^3C_2=\dfrac{3\times2}{2!}=3$ .

Answer

6\text{ arrangements},\quad3\text{ selections}

Worked example

Same pool, different question

Arrange 3 digits from 7 digits: ${}^7P_3=7\times6\times5=210$ .
Choose 3 fish from 7 fish: ${}^7C_3=\dfrac{7\times6\times5}{3!}=35$ .

Answer

210\text{ arrangements},\quad35\text{ selections}

For several independent choices, count each part and multiply. Choosing 2 boys from 6 and 1 girl from 4 gives ${}^6C_2\times{}^4C_1=60$ .
For at least or at most, list the legal splits first. The formula comes after the split is clear.

Worked example

Small case split

Choose 3 musicians from 5 violinists and 4 guitarists. The group needs at least 2 violinists.

Legal split 1: 2 violinists and 1 guitarist: ${}^5C_2\,{}^4C_1=40$ .
Legal split 2: 3 violinists and 0 guitarists: ${}^5C_3=10$ .
Add the legal cases because they cannot happen at the same time: $40+10=50$ .

Answer

50

Exam version of the same split

The real-paper version below is the same idea as the small split: write every legal split, count each split, then add.
Choose 5 students from 10 violinists, 6 guitarists and 5 pianists. The group needs at least 2 violinists, at least 1 guitarist and at most 1 pianist. (9709/52/O/N/24 Q5(b))

Split (V,G,P)	Ways
(4,1,0)	${}^{10}C_4\,{}^6C_1=1260$
(3,2,0)	${}^{10}C_3\,{}^6C_2=1800$
(2,3,0)	${}^{10}C_2\,{}^6C_3=900$
(3,1,1)	${}^{10}C_3\,{}^6C_1\,{}^5C_1=3600$
(2,2,1)	${}^{10}C_2\,{}^6C_2\,{}^5C_1=3375$

Answer

1260+1800+900+3600+3375=10\,935

Common mistake

Do not add a stray

\times r!

to a team or group question. The group is already chosen; its order is ignored.

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Permutations vs combinations

This is the main decision in the chapter: does order matter?
If order matters, use $P$ . If order does not matter, use $C$ .
A combination is a permutation with the extra orderings divided out.

P arranges; C chooses

Start with one question: would swapping the chosen objects make a new answer?
If swapping changes the answer, order matters, so use $P$ . If swapping does not change the answer, order is ignored, so use $C$ .
Case splits are a later step. First learn the small difference between $AB$ and $BA$ .

${}^nC_r=\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!(n-r)!}$ . The $r!$ removes the order of the chosen group.

Question wording	Order?	Use
arrange, code, row, queue, seats, podium	yes	${}^nP_r$
choose, select, team, committee, hand, group	no	${}^nC_r$
several independent choices	depends on each choice	multiply the counts
at least / at most	case split	add legal cases only

choose: ⁿCr = 10×r! = 6=arrange: ⁿPr = 60

Worked example

First rung: three letters

Arrange 2 letters from A, B and C. AB and BA are different, so order matters: ${}^3P_2=3\times2=6$ .
Choose 2 letters from A, B and C. AB and BA are the same pair, so order does not matter: ${}^3C_2=\dfrac{3\times2}{2!}=3$ .

Answer

6\text{ arrangements},\quad3\text{ selections}

Worked example

Same pool, different question

Arrange 3 digits from 7 digits: ${}^7P_3=7\times6\times5=210$ .
Choose 3 fish from 7 fish: ${}^7C_3=\dfrac{7\times6\times5}{3!}=35$ .

Answer

210\text{ arrangements},\quad35\text{ selections}

For several independent choices, count each part and multiply. Choosing 2 boys from 6 and 1 girl from 4 gives ${}^6C_2\times{}^4C_1=60$ .
For at least or at most, list the legal splits first. The formula comes after the split is clear.

Worked example

Small case split

Choose 3 musicians from 5 violinists and 4 guitarists. The group needs at least 2 violinists.

Legal split 1: 2 violinists and 1 guitarist: ${}^5C_2\,{}^4C_1=40$ .
Legal split 2: 3 violinists and 0 guitarists: ${}^5C_3=10$ .
Add the legal cases because they cannot happen at the same time: $40+10=50$ .

Answer

50

Exam version of the same split

The real-paper version below is the same idea as the small split: write every legal split, count each split, then add.
Choose 5 students from 10 violinists, 6 guitarists and 5 pianists. The group needs at least 2 violinists, at least 1 guitarist and at most 1 pianist. (9709/52/O/N/24 Q5(b))

Split (V,G,P)	Ways
(4,1,0)	${}^{10}C_4\,{}^6C_1=1260$
(3,2,0)	${}^{10}C_3\,{}^6C_2=1800$
(2,3,0)	${}^{10}C_2\,{}^6C_3=900$
(3,1,1)	${}^{10}C_3\,{}^6C_1\,{}^5C_1=3600$
(2,2,1)	${}^{10}C_2\,{}^6C_2\,{}^5C_1=3375$

Answer

1260+1800+900+3600+3375=10\,935

Common mistake

Do not add a stray

\times r!

to a team or group question. The group is already chosen; its order is ignored.