Enumeration – OCR FSMQ Additional Maths

Enumeration means counting without listing. Faced with "how many ways…?", the aim is to find the answer by multiplying and dividing, not by writing out every possibility.

This section is the newest part of the course for most students. It combines the binomial expansion with permutations, combinations and their use in probability — which is why the same numbers, $\binom{n}{r}$, turn up in all of them.

The single most useful fact in this topic

The numbers in Pascal's triangle, the coefficients in $(a+b)^n$,
and the number of ways of choosing $r$ things from $n$
are all the same numbers: $\;{}^nC_r$

Your calculator has these built in. Look for the nCr and nPr buttons. The specification explicitly names both notations, so you are expected to use them.

Word	Meaning	Example
Factorial	$n! = n \times (n-1) \times \cdots \times 2 \times 1$	$5! = 120$
Arrangement	An ordering of objects	ABC and BAC are different
Permutation	An ordered selection of $r$ from $n$	Which two win gold and silver
Combination	An unordered selection of $r$ from $n$	Which two sit on a committee
Product rule	Independent choices multiply	$3$ starters $\times$ $4$ mains $= 12$ meals
Binomial	An expression with two terms	$a + b$, $2x - 3$
Trial	One repetition of an experiment	One roll of a die

The notations, all meaning the same thing

${}^nC_r \;=\; \binom{n}{r} \;=\; \dfrac{n!}{r!\,(n-r)!}$

Order is the whole question. Before you compute anything, ask: does swapping two chosen items give a different outcome? If yes, it is a permutation. If no, a combination. Getting this wrong is the single biggest source of lost marks in this topic.

Product rule (EN3)

$m$ ways then $n$ ways $\Rightarrow$ $m \times n$ ways altogether

Arrangements of $n$ distinct objects

$n!$

Permutations — ordered, $r$ from $n$ (EN4)

${}^nP_r = \dfrac{n!}{(n-r)!}$

Combinations — unordered, $r$ from $n$ (EN5)

${}^nC_r = \dfrac{n!}{r!\,(n-r)!} = \dfrac{{}^nP_r}{r!}$

Binomial expansion (EN1)

$(a+b)^n = \displaystyle\sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^{r}$

Binomial probability (EN2, EN6)

$P(X = r) = \binom{n}{r} p^r (1-p)^{n-r}$

Worked Example — Telling the four apart

Ten athletes run a race. Count the ways for each of these.

①Full finishing order: all ten in order $= 10! = 3\,628\,800$

②Gold, silver, bronze: ordered, $3$ from $10$ $= {}^{10}P_3 = 10 \times 9 \times 8 = 720$

③A team of three to go forward: unordered $= {}^{10}C_3 = \dfrac{720}{6} = 120$

④Each athlete passes or fails a drug test: two options each, so $2^{10} = 1024$ possible sets of results.

Notice ② and ③ differ by exactly $3! = 6$ — the number of ways of ordering the three chosen athletes. That relationship, ${}^nC_r = {}^nP_r \div r!$, is worth remembering.

EN1The Binomial ExpansionPascal's triangle, $\binom{n}{r}$, and expanding $(a+b)^n$ for positive integer $n$.
EN3The Product Rule for CountingMultiplying choices, arrangements, and $n!$.
EN4PermutationsOrdered selections, and the ${}^nP_r$ notation.
EN5CombinationsUnordered selections, and the ${}^nC_r$ notation.
EN2Representing OutcomesTree diagrams, two-way tables and Venn diagrams for enumerating outcomes.
EN6Counting in ProbabilityThe binomial distribution, and "at least" problems.

Using ${}^nP_r$ when order does not matter. A committee of $3$ from $10$ is ${}^{10}C_3 = 120$, not ${}^{10}P_3 = 720$. You would be counting each committee $6$ times.

Adding when you should multiply. Independent stages multiply. Only add when you are counting separate, non-overlapping cases.

Forgetting that $\binom{n}{r}$ counts, so $r$ must be a whole number with $0 \leqslant r \leqslant n$. If your working produces $\binom{5}{7}$, you have made a slip.

Mishandling "at least". $P(\text{at least one}) = 1 - P(\text{none})$. Adding up every case from one upwards is slow and error-prone.

Dropping the coefficient in a binomial term. In $(2x+3)^5$, the term in $x^2$ is $\binom{5}{2}(2x)^2 3^3$ — the $2$ gets squared and the $3$ gets cubed. Both are easy to forget.

First question

Does order matter?

Product rule

Stages multiply.

$n!$

Arrangements of all $n$ objects.

${}^nP_r$

Ordered: $\dfrac{n!}{(n-r)!}$.

${}^nC_r$

Unordered: $\dfrac{n!}{r!(n-r)!}$.

Link

${}^nC_r = {}^nP_r \div r!$.

Symmetry

$\binom{n}{r} = \binom{n}{n-r}$.

Binomial term

$\binom{n}{r}a^{n-r}b^r$.

Binomial probability

$\binom{n}{r}p^r(1-p)^{n-r}$.

At least one

$1 - P(\text{none})$.

Question 1

Evaluate $6!$ and ${}^6C_2$.

▶ Show solution

$6! = 720$

${}^6C_2 = \dfrac{6 \times 5}{2 \times 1} = 15$

Question 2

A menu has $4$ starters, $6$ mains and $3$ desserts. How many three-course meals are possible?

▶ Show solution

Independent choices, so multiply:

$4 \times 6 \times 3 = 72$ meals.

Question 3

In how many orders can $7$ books be arranged on a shelf?

▶ Show solution

All $7$ arranged, so $7! = 5040$.

Question 4

From a squad of $15$, how many different teams of $11$ can be picked?

▶ Show solution

Order does not matter, so ${}^{15}C_{11}$.

Using the symmetry $\binom{15}{11} = \binom{15}{4} = \dfrac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1} = 1365$

Question 5

Write out the expansion of $(1 + x)^4$.

▶ Show solution

Row $4$ of Pascal's triangle is $1, 4, 6, 4, 1$.

$1 + 4x + 6x^2 + 4x^3 + x^4$

Question 6

Three prizes — first, second and third — are awarded among $8$ entrants. In how many ways?

▶ Show solution

The prizes are different, so order matters:

${}^8P_3 = 8 \times 7 \times 6 = 336$

Question 7

A fair coin is tossed $6$ times. Find the probability of exactly $4$ heads.

▶ Show solution

$P(X=4) = \binom{6}{4}(0.5)^4(0.5)^2 = 15 \times (0.5)^6$

$= \dfrac{15}{64} = 0.234$ (3 s.f.)

Question 8

Find the term in $x^3$ in the expansion of $(2 + x)^5$.

▶ Show solution

The term is $\binom{5}{3} 2^{2} x^{3}$

$= 10 \times 4 \times x^3 = 40x^3$

Question 9

Five dice are rolled. Find the probability of at least one six.

▶ Show solution

Use the complement. $P(\text{no six on one die}) = \tfrac56$.

$P(\text{no sixes at all}) = \left(\tfrac56\right)^5 = \dfrac{3125}{7776}$

$P(\text{at least one}) = 1 - \dfrac{3125}{7776} = \dfrac{4651}{7776} = 0.598$ (3 s.f.)

Question 10

A committee of $4$ is chosen from $6$ women and $5$ men.

(a) How many committees are possible altogether? (b) How many contain exactly $2$ women? (c) How many contain at least $1$ man? (d) A chairperson and a secretary are then chosen from the committee of $4$. How many ways?

▶ Show solution

(a) Choosing $4$ from $11$, order irrelevant:

${}^{11}C_4 = \dfrac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330$

(b) Choose $2$ women from $6$ and $2$ men from $5$ — two independent stages, so multiply:

${}^6C_2 \times {}^5C_2 = 15 \times 10 = 150$

(c) "At least one man" is easiest as the complement of "no men", i.e. all four women:

${}^6C_4 = 15$ all-women committees.

$330 - 15 = 315$

(d) Now order does matter, because chair and secretary are different roles:

${}^4P_2 = 4 \times 3 = 12$ ways.

This last part is the point of the whole topic in miniature: the same four people give $1$ committee but $12$ officer pairs, purely because order matters in one case and not the other.