H2 Math Permutations and Combinations

JC2 · H2 Mathematics · Study guide

Choose and arrange under restrictions, with complete working for repeated objects, distributions, groups, blocks, gaps and rotational symmetry.

Concepts · worked methods · graphs10 min read · practice linked below

What is being counted?

Before choosing a formula, decide whether the objects are distinct or identical, whether order matters, and which arrangements count as the same. “Choose a team” and “assign its captain and secretary” describe different outcomes. Multiply for consecutive choices; add only mutually exclusive cases.

Choose first, then arrange if needed

Select four objects with a colour restriction

A box contains 5 distinct red tokens, 4 distinct blue tokens and 3 distinct green tokens. Four tokens are selected, with order irrelevant. Find the number of selections with at least two colours, and with all three colours.

For at least two colours, subtract selections that use only one colour. Green has fewer than four tokens, so contributes no such selection.

\begin{aligned}N&=\binom{12}{4}-\binom54-\binom44\\&=495-5-1\\&=489.\end{aligned}

With all three colours, exactly one colour occurs twice. These cases do not overlap:

\begin{aligned}N&=\binom52\binom41\binom31\\&\quad+\binom51\binom42\binom31\\&\quad+\binom51\binom41\binom32\\&=120+90+60\\&=270.\end{aligned}

There is no extra factor of $4!$ : a selection does not record the order in which tokens are drawn.

Repeated letters and repeated digits

Arrange repeated letters, then form shorter words

How many arrangements of all the letters of BANANA are there? How many four-letter codewords can be made from its available letters?

\begin{aligned}N_{\text{all}}&=\frac{6!}{3!2!}\\&=60.\end{aligned}

For shorter codewords, dividing $6!$ does not work: first decide which multiplicities occur. The available letters are A three times, N twice and B once.

Multiplicity	Choice and arrangement	Count
3 + 1	AAA with N or B: $2(4!/3!)$	8
2 + 2	AANN: $4!/(2!2!)$	6
2 + 1 + 1	repeat A or N, use the other and B: $2(4!/2!)$	24

\begin{aligned}N_{\text{four-letter}}&=8+6+24\\&=38.\end{aligned}

The letter B cannot be repeated. Every valid codeword appears in exactly one row.

A six-digit number with exactly three even digits

Digits 1–9 may be repeated. Find the number of six-digit numbers containing exactly three even digits. Then count six-digit odd numbers below 600000 when repetition is forbidden.

For the first task, choose the three even positions, then fill them. There are four even digits and five odd digits:

\begin{aligned}N&=\binom63 4^3 5^3\\&=160000.\end{aligned}

For the second task, the last digit must be odd and the first digit is 1–5. Split on whether the first digit is odd or even, because that changes the last-digit choices.

\begin{aligned}N&=3(4)\,{}^7P_4+2(5)\,{}^7P_4\\&=(12+10)\frac{7!}{3!}\\&=18480.\end{aligned}

After choosing the first and last digits, seven unused digits remain for four ordered middle positions.

Are the groups labelled?

Three teams of equal size

Nine distinct people are split into three teams of three. Find the number of splits if the teams are unlabelled, then if the teams play three different named games. For unlabelled teams, count the splits in which two specified people are together.

\begin{aligned}N_{\text{unlabelled}}&=\frac{\binom93\binom63\binom33}{3!}\\&=280,\\N_{\text{named}}&=\binom93\binom63\binom33\\&=1680.\end{aligned}

Divide by $3!$ only in the first task: permuting the same three teams creates no new split. For the specified pair, form their team first, then split the remaining six people into two unlabelled teams:

\begin{aligned}N_{\text{together}}&=\binom71\frac{\binom63}{2!}\\&=70.\end{aligned}

Check: the probability that a particular other person joins the first person is $2/8=1/4$ ; $70/280=1/4$ , as expected.

Distribute objects: identify what can be distinguished

Five distinct gifts, at least one each

Five different gifts are given to three named people, each receiving at least one. Count by the size of each person’s share.

\begin{aligned}N_{(3,1,1)}&=3\binom53 2!\\&=60,\\N_{(2,2,1)}&=3\binom52\binom32\\&=90,\\N&=60+90=150.\end{aligned}

In the second line choose the person receiving one gift, then give two gifts to the first of the other two named people and two to the other. Their names already distinguish them; no further $2!$ is needed.

If all five gifts were identical, only the share sizes matter. The positive splits are permutations of $(3,1,1)$ and $(2,2,1)$ : $3+3=6$ distributions.

“Either … or … or both”: subtract the overlap

A row uses 4 identical E pieces, 3 identical C pieces and 2 identical W pieces. Count rows whose first piece is E or whose last two pieces are E, or both.

Let A be “first is E” and B be “last two are E”. Fix those positions before arranging the remaining repeated pieces:

\begin{aligned}n(A)&=\frac{8!}{3!3!2!}=560,\\n(B)&=\frac{7!}{2!3!2!}=210,\\n(A\cap B)&=\frac{6!}{3!2!}=60,\\n(A\cup B)&=560+210-60\\&=710.\end{aligned}

In the overlap, three E pieces are fixed, leaving one E. Do not subtract an arrangement in which the required E positions are not all fixed.

Blocks, gaps and “exactly”

Exactly three of four friends sit together

Four specified friends and three other distinct people stand in a row. Count arrangements in which there is a block of exactly three friends, with the fourth friend separated from that block.

Choose the isolated friend. Arrange the block of three internally, then arrange this block with the three other people. Those four units create five gaps. The two gaps touching the block are forbidden to the isolated friend, leaving three.

\begin{aligned}N&=4\times3!\times4!\times3\\&=1728.\end{aligned}

The block of three is unique in each valid row, so nothing is counted twice. Treating three friends as a block without keeping the fourth away would include all four together.

Separate two different pairs of repeated letters

Arrange A, A, B, B, C, D in a row so that no identical letters are adjacent. Use complementation: a gap method that first separates A may let later B insertions create new choices inconsistently.

\begin{aligned}N_{\text{total}}&=\frac{6!}{2!2!}=180,\\N_{AA}&=\frac{5!}{2!}=60,\\N_{BB}&=\frac{5!}{2!}=60,\\N_{AA,BB}&=4!=24,\\N_{\text{required}}&=180-60-60+24\\&=84.\end{aligned}

For $N_{AA,BB}$ , the four units are AA, BB, C and D. Add this overlap back because the first two subtractions removed it twice.

A selection of any non-zero size

Each person is chosen or not chosen

A non-empty delegation is chosen from eight people, with two specified people either both included or both excluded. The pair has two choices; each of the other six people has two choices.

\begin{aligned}N_{\text{including empty}}&=2\times2^6\\&=128,\\N_{\text{non-empty}}&=128-1=127.\end{aligned}

Subtract the empty delegation once. Summing combinations by delegation size is possible, but the independent yes/no choices are shorter here.

Around a circle: arrange the blocks, then use the gaps

Four pairs and two separated hosts

Four teams of two distinct people and two distinct hosts sit around a round table. Teammates must sit together and the hosts must not sit next to each other. Rotations count as the same; reflections count as different.

Arrange the four team blocks around the circle, then arrange each team internally. Put the hosts in two different gaps between team blocks:

\begin{aligned}N&=(4-1)!(2!)^4\binom42 2!\\&=6(16)(6)(2)\\&=1152.\end{aligned}

There are four gaps in a circle. Choosing different gaps ensures that a team block separates the hosts. The internal order of each pair matters.

Choose the hosts’ gaps

Each highlighted pair of gaps supports two orders of the distinct hosts.

Selected gaps

Gaps 1 and 2: host H can use either selected gap, with host J in the other. Six gap pairs × two orders = 12 placements.

Variation: seats on a square table

Eight distinct people sit two per side of a square table. Rotations that carry the whole seating layout onto itself count as the same. Reflections remain different.

\begin{aligned}N&=\frac{8!}{4}\\&=10080.\end{aligned}

There are four valid rotations, not eight: rotating by one seat does not preserve the square. If a door fixes an absolute orientation, all eight seats are distinguishable and the answer is $8!$ . With four boys and four girls, one of each on every side:

\begin{aligned}N&=\frac{2^4\,4!\,4!}{4}\\&=2304.\end{aligned}

Choose which seat on each side is occupied by a boy, then fill the boy and girl positions separately, before dividing by the four rotations.

Before you finish

Check that cases are exhaustive and do not overlap, and that each factorial has a reason. “Not all together” removes one block case; “none together” requires separation. Divide by a group factorial only when exchanging entire groups creates the same outcome.