Arrangements and selections · Counting

When the list of outcomes is too long to write out, we count it instead. This topic comes up mostly in JAMB, and it’s the counting behind many probability questions.

The multiplication rule

If one choice can be made in $a$ ways and, after it, a second choice in $b$ ways, the two together can be made in $a \times b$ ways.

A PIN has 3 digits, each 0 to 9: $10 \times 10 \times 10 = 1000$ PINs. If the digits must be different: $10 \times 9 \times 8 = 720$ , because each digit used is no longer available.

One box per choiceMultiply the number of choices for each place

More: the multiplication rule

JAMB 2002 · UME · Q28How many three-digit numbers can be formed from 32564 without any digit being repeated?

Factorials

Arranging $n$ different things in a row: $n$ choices for the first place, $n - 1$ for the second, and so on down to 1. That product is $n$ factorial:

n! = n \times (n - 1) \times \cdots \times 2 \times 1

So 5 people can stand in a row in $5! = 120$ ways. By agreement, $0! = 1$ .

Arrangements: order matters

The number of ways to arrange $r$ things chosen from $n$ different things is

{}^nP_r = n \times (n - 1) \times \cdots \ (r \text{ factors}) = \frac{n!}{(n - r)!}

Use it when swapping two chosen things gives a different result: first and second prize, a president and a secretary, seats in a row, digits in a number.

More: arrangements

JAMB 2013 · UTME · Q48In how many ways can 3 seats be occupied if 5 people are willing to sit?

Selections: order doesn’t matter

The number of ways to choose $r$ things from $n$ , when only which things are chosen matters, is

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

Use it for a committee, a team, a hand of cards, subjects for an exam: choosing Ade and Bola is the same as choosing Bola and Ade.

Why ⁿPᵣ = ⁿCᵣ × r!

Arrange or select?Change n and r

ABABBA

ACACCA

ADADDA

BCBCCB

BDBDDB

CDCDDC

6selections, ⁿCᵣ = 4! ÷ (2! × 2!)2orders of each, 2!12arrangements, ⁿPᵣ = 4! ÷ 2!

n (letters) = 4r (chosen) = 2

Order matters (a president and a secretary, a PIN, a word): each selection gives 2 different arrangements, so 6 × 2 = 12. That is ⁿPᵣ = ⁿCᵣ × r!.

Selections and arrangementsEach selection of r things gives r! orders

Each row is one selection. With Order matters chosen, the row unfolds into all the ways of ordering it, and every selection unfolds into the same number of arrangements, $r!$ . So

{}^nP_r = {}^nC_r \times r! \qquad\text{and}\qquad {}^nC_r = \frac{{}^nP_r}{r!}

More: selections

Committees with conditions

When a committee needs so many from one group and so many from another, choose from each group separately and multiply.

Worked example · JAMB 2010

JAMB 2010 · UTME · Q49

In how many ways can a committee of 2 women and 3 men be chosen from 6 men and 5 women?

The women
${}^5C_2 = \frac{5 \times 4}{2} = 10$ .
Think first. Choose 2 women from 5: how many ways?
The men
${}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ .
Think first. Choose 3 men from 6: how many ways?
Together
Any choice of women goes with any choice of men: $10 \times 20 = 200$ . (B)

For “at least one man and at least one woman”, either add the separate cases (1 man and 2 women, 2 men and 1 woman), or take the total and subtract the cases that break the rule (all men, all women).

More: committees with conditions

Letters that repeat

Arranging the letters of a word where some letters are the same: swapping two identical letters gives the same word, so divide by the factorial of each repeat.

\text{arrangements} = \frac{n!}{p!\,q!\cdots}

where the letters repeat $p$ times, $q$ times, ….

Worked example · JAMB 2012

JAMB 2012 · UTME · Q48

In how many ways can the letters of the word TOTALITY be arranged?

Count the letters
T, O, T, A, L, I, T, Y: 8 letters, and T appears 3 times. The others appear once.
Think first. How many letters? Which ones repeat?
Divide out the repeats
$\frac{8!}{3!} = \frac{40\,320}{6} = 6720$
(A)

More: letters that repeat

Round a table

For $n$ people round a circular table, only their positions relative to each other matter. Fix one person’s seat and arrange the rest: $(n - 1)!$ . Six people: $5! = 120$ ways.

Round a tableTurning the table gives the same seating

More: round a table

JAMB 2011 · UTME · Q48In how many ways can five people sit round a circular table?

Your turn

JAMB 1992 · UME · Q49

Counting

Five people are to be arranged in a row for a group photograph. How many arrangements are there if a married couple in the group insist on sitting next to each other?

Worked solution (try it first)

Tie the couple together and treat them as one unit.
Then there are 4 units to arrange in a row: $4! = 24$ ways.
Inside the unit the couple can swap places: $2! = 2$ ways.
Multiply: $24 \times 2 = 48$ arrangements, option A.

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