5.2.3 Combinations - GCEALevelMaths9709

In this topic we will learn how to:

understand and solve problems involving combinations

A combination refers to an arrangement in which the order does not matter. The notation for combinations is,
$^{n}C_{r}$ You should have come across it in binomial expansion under the topic Series, in pure mathematics $1$ . The above reads as $n$ choose $r$ , where $n$ represents the number of choices and $r$ represents the selections or slots.

Let’s walk through past paper questions.

1. A group of $6$ people is to be chosen from $4$ men and $11$ women. (9709/52/O/N/21 number 2)

(a) In how many different ways can a group of $6$ be chosen if it must contain exactly $1$ man?

If the group must have exactly $1$ man, it means we must have exactly $5$ women for the group to have a total of $6$ people. For the men we are choosing $1$ out of $4$ men,
$^{4}C_{1}$ For the women we are choosing $5$ out of $11$ women,
$^{11}C_{5}$ Putting everything together,
$^{4}C_{1} \times ^{11}C_{5}$ Which simplifies to give,
$4 \times 462$ $1\ 848$ Therefore, the final answer is,
$1\ 848$ Two of the $11$ women are sisters Jane and Kate.
(b) In how many different ways can a group of $6$ be chosen if Jane and Kate cannot both be in the group?

Let’s assign Jane a letter of $J$ and Kate a letter of $K$ . The problem, is asking us to evaluate the following three scenarios,
$\textmd{\textcolor{#2192ff}{Jane is in the group and Kate is not}}$ $\textmd{\textcolor{#0f0}{Kate is in the group and Jane is not}}$ $\textmd{\textcolor{red}{Neither of the two are in the group}}$ Let’s evaluate the first scenario, Jane is in the group and Kate is not. It would look something like this,
$\textmd{4 men and 9 women}$ $\textmd{13 people}$ $\underline{J}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $9$ women because Jane has already been selected, so we remove her from the pool of selection. Kate is also removed from the pool of selection to ensure that we do not select her.

Since there are $13$ choices and only $5$ slots left, this becomes,
$^{13}C_{5}$ Which simplifies to,
$\textcolor{#2192ff}{1\ 287}$ Let’s evaluate the second scenario, Kate is in the group and Jane is not. It would look something like this,
$\textmd{4 men and 9 women}$ $\textmd{13 people}$ $\underline{K}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $9$ women because Kate has already been selected, so we remove her from the pool of selection. Jane is also removed from the pool of selection to ensure that we do not select her.

Since there are $13$ choices and only $5$ slots left, this becomes,
$^{13}C_{5}$ Which simplifies to,
$\textcolor{#0f0}{1\ 287}$ Let’s evaluate the third scenario, neither of Jane and Kate are in the group. It would look something like this,
$\textmd{4 men and 9 women}$ $\textmd{13 people}$ $\underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $9$ women because both Jane and Kate have been removed to ensure that we do not select them.

Since there are $13$ choices and $6$ slots left, this becomes,
$^{13}C_{6}$ Which simplifies to,
$\textcolor{red}{1\ 716}$ Putting everything together,
$\textmd{Total } = \textcolor{#2192ff}{1\ 287} + \textcolor{#0f0}{1\ 287} + \textcolor{red}{1\ 716}$ $\textmd{Total } = 4\ 290$ Therefore, the final answer is,
$4\ 290$ 2. There are $6$ men and $8$ women in a Book Club. The committee of the club consists of five of its members. Mr Lan and Mrs Lan are members of the club. (9709/51/M/J/22 number 2)

(a) In how many different ways can the committee be selected if exactly one of Mr Lan and Mrs Lan must be on the committee?

To solve this problem we have to consider the following two scenarios,
$\textmd{\textcolor{#2192ff}{Mr Lan is on the committee and Mrs Lan is not}}$ $\textmd{\textcolor{#0f0}{Mrs Lan is on the committee and Mr Lan is not}}$ Let’s evaluate the first scenario, Mr Lan is on the committee and Mrs Lan is not. It would look something like this,
$\textmd{5 men and 7 women}$ $\textmd{12 people}$ $\underline{Mr}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are $5$ men because Mr Lan has already been selected so we remove him from the pool of selection. There are $7$ women because Mrs Lan has been removed from the pool of selection to ensure that we do not pick her.

Since there are $12$ choices and $4$ slots remaining, it becomes,
$^{12}C_{4}$ Which simplifies to give,
$\textcolor{#2192ff}{495}$ Let’s evaluate the second scenario, Mrs Lan is on the committee and Mr Lan is not. It would look something like this,
$\textmd{5 men and 7 women}$ $\textmd{12 people}$ $\underline{Mrs}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are $5$ men because Mr Lan has been removed from the pool of selection to ensure that we do not pick him. There are $7$ women because Mrs Lan has already been selected so we remove her from the pool of selection.

Since there are $12$ choices and $4$ slots remaining, it becomes,
$^{12}C_{4}$ Which simplifies to give,
$\textcolor{#0f0}{495}$ Putting everything together,
$\textmd{Total } = \textcolor{#0f0}{495} + \textcolor{#2192ff}{495}$ $\textmd{Total } = 990$ Therefore, the final answer is,
$990$ (b) In how many different ways can the committee be selected if Mrs Lan must be on the committee and there must be more women than men on the committee?

To solve this problem there are three scenarios we must consider,
$\textcolor{#2192ff}{4\textmd{ women and }0\textmd{ men}}$ $\textcolor{#0f0}{3\textmd{ women and }1\textmd{ man}}$ $\textcolor{red}{2\textmd{ women and }2\textmd{ men}}$ Let’s evaluate the first scenario, $4$ women and $0$ men. It would look something like this,
$\textmd{6 men and 7 women}$ $\underline{Mrs}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $7$ women because Mrs Lan has already been selected, so we remove her from the pool of selection.

We want to select $4$ women from $7$ and $0$ men from $6$ ,
$^{7}C_{4} \times ^{6}C_{0}$ Which simplifies to give,
$35 \times 1$ $\textcolor{#2192ff}{35}$ Let’s evaluate the second scenario, $3$ women and $1$ man. It would look something like this,
$\textmd{6 men and 7 women}$ $\underline{Mrs}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $7$ women because Mrs Lan has already been selected, so we remove her from the pool of selection.

We want to select $3$ women from $7$ and $1$ man from $6$ ,
$^{7}C_{3} \times ^{6}C_{1}$ Which simplifies to give,
$35 \times 6$ $\textcolor{#0f0}{210}$ Let’s evaluate the third scenario, $2$ women and $2$ men. It would look something like this,
$\textmd{6 men and 7 women}$ $\underline{Mrs}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}\ \ \underline{\textcolor{transparent}{a}}$ Note: There are only $7$ women because Mrs Lan has already been selected, so we remove her from the pool of selection.

We want to select $2$ women from $7$ and $2$ men from $6$ ,
$^{7}C_{2} \times ^{6}C_{2}$ Which simplifies to give,
$21 \times 15$ $\textcolor{red}{315}$ Putting everything together,
$\textmd{Total } = \textcolor{#2192ff}{35} + \textcolor{#0f0}{210} + \textcolor{red}{315}$ $\textmd{Total } = 560$ Therefore, the final answer is,
$560$ 3. Donna has $2$ necklaces, $8$ rings and $4$ bracelets, all different. She chooses $4$ pieces of jewellery. How many possible selections can she make if she chooses at least $1$ necklace and at least $1$ bracelet? (9709/62/M/J/18 number 6)

To be able to select at least $1$ necklace and at least $1$ bracelet and select only four pieces of jewellery we have to consider the scenarios below,
$1\textmd{ necklace and }1\textmd{ bracelet and }2\textmd{ rings}$ $1\textmd{ necklace and }2\textmd{ bracelets and }1\textmd{ ring}$ $1\textmd{ necklace and }3\textmd{ bracelets and }0\textmd{ rings}$ $2\textmd{ necklaces and }1\textmd{ bracelet and }1\textmd{ ring}$ $2\textmd{ necklaces and }2\textmd{ bracelets and }0\textmd{ rings}$ Note: We cannot pick more than $2$ necklaces, because Donna only has $2$ necklaces.

Let’s evaluate the first scenario, $1$ necklace and $1$ bracelet and $2$ rings. Donna picks $1$ necklace from $2$ , $1$ bracelet from $4$ and $2$ rings from $8$ ,
$^{2}C_{1} \times ^{4}C_{1} \times ^{8}C_{2}$ Which simplifies to give,
$2 \times 4 \times 28$ $224$ Let’s evaluate the second scenario, $1$ necklace and $2$ bracelets and $1$ ring. Donna picks $1$ necklace from $2$ , $2$ bracelets from $4$ and $1$ ring from $8$ ,
$^{2}C_{1} \times ^{4}C_{2} \times ^{8}C_{1}$ Which simplifies to give,
$2 \times 6 \times 8$ $96$ Let’s evaluate the third scenario, $1$ necklace and $3$ bracelets and $0$ rings. Donna picks $1$ necklace from $2$ , $3$ bracelet from $4$ and $0$ rings from $8$ ,
$^{2}C_{1} \times ^{4}C_{3} \times ^{8}C_{0}$ Which simplifies to give,
$2 \times 4 \times 1$ $8$ Let’s evaluate the fourth scenario, $2$ necklaces and $1$ bracelet and $1$ ring. Donna picks $2$ necklaces from $2$ , $1$ bracelet from $4$ and $1$ ring from $8$ ,
$^{2}C_{2} \times ^{4}C_{1} \times ^{8}C_{1}$ Which simplifies to give,
$1 \times 4 \times 8$ $32$ Let’s evaluate the fifth scenario, $2$ necklaces and $2$ bracelets and $0$ rings. Donna picks $2$ necklaces from $2$ , $2$ bracelets from $4$ and $0$ rings from $8$ ,
$^{2}C_{2} \times ^{4}C_{2} \times ^{8}C_{0}$ Which simplifies to give,
$1 \times 6 \times 1$ $6$ Putting everything together,
$\textmd{Total } = 224 + 96 + 8 + 32 + 6$ $\textmd{Total } = 366$ Therefore, the final answer is,
$366$