Permutations and Combinations

Lesson

So we have just been looking at the topic of combinations. A world where order doesn't matter and we are counting the number of combinations. We have learnt the new notation ^{n}C_{r} and C(n,r), and the formula $\frac{n!}{r!(n-r!)}$`n`!`r`!(`n`−`r`!).

This set just looks at slightly more difficult questions and questions where we impose some restrictions on the combinations.

A selection of $4$4 people are to be chosen from a group of $9$9 people. How many selections are possible if the youngest or oldest is included but not both?

$4$4 people can be chosen from $9$9 in ^{9}C_{4} ways, which is $\frac{9!}{4!5!}=126$9!4!5!=126 ways.

Make sure you know how to calculate combination on your calculator using the relevant button or command.

BUT, we have a restriction regarding the oldest or youngest.

So what we have here is two possible situations to consider.

In each case, one person must definitely fill one of the $4$4 positions, this leaves $3$3 positions to fill. Both youngest and oldest get removed from the $9$9 people since their selections are already known. This means that there will be $7$7 people left to choose from.

This results in

$C(7,3)+C(7,3)=2\times C(7,3)=2\times\frac{7!}{3!4!}=2\times\frac{210}{6}=2\times35=70$`C`(7,3)+`C`(7,3)=2×`C`(7,3)=2×7!3!4!=2×2106=2×35=70.

So there are $70$70 ways this selection can be made.

Sometimes it helps to have a method that keeps track of the whole situation.

- The first thing to occur, is that we choose the oldest, we do that in
^{1}C_{1}ways. - The second thing is then to ignore the youngest, we do that in
^{1}C_{0}ways - The third thing is then to choose the remaining $3$3, and we do that in
^{7}C_{3}ways.

Another notation often used for combinations looks like this

$\binom{n}{r}=C(n,r)=$(`n``r`)=`C`(`n`,`r`)= ^{n}C_{r} so our three steps now become

- The first thing to occur, is that we choose the oldest, we do that in $\binom{1}{1}$(11) ways.
- The second thing is then to ignore the youngest, we do that in $\binom{1}{0}$(10) ways
- The third thing is then to choose the remaining $3$3, and we do that in $\binom{7}{3}$(73) ways.

When we write these as a multiplication, we get

$\binom{1}{1}\binom{1}{0}\binom{7}{3}$(11)(10)(73)

Look at the top row here (the $n$`n` positions).... $1+1+7=9$1+1+7=9 which is the total number of people we had to choose from

Look at the bottom row here (the $r$`r` positions).... $1+0+3=4$1+0+3=4 which is the total number of people we were choosing.

This method is great to keep track of everyone. We don't want to leave anyone out!

So to complete this question....

$\binom{1}{1}\binom{1}{0}\binom{7}{3}+\binom{1}{1}\binom{1}{0}\binom{7}{3}=2\times\binom{1}{1}\binom{1}{0}\binom{7}{3}=1\times1\times70=70$(11)(10)(73)+(11)(10)(73)=2×(11)(10)(73)=1×1×70=70

A team of $3$3 is to be chosen at random from a group of $5$5 girls and $6$6 boys.

In how many ways can the team be chosen if:

there are no restrictions?

there must be more boys than girls?

$5$5 boys and $6$6 girls are part of the debate team. $4$4 of them must represent the school at the upcoming debate tournament. In how many ways can the team of $4$4 be formed if:

it must contain $2$2 boys and $2$2 girls?

it must contain at least $1$1 boy and $1$1 girl?

Use permutations and combinations