Question

How many $4$ person committees are possible from a group of $9$ people if
A) There are no restrictions?
B) Both John and Barbara must be on the committee?
C) Either John or Barbara (but not both) must be on the committee?

Answer 1

To solve this question, we need to use the concept of combinations of $r$ objects out of a total of $n$ objects. We have to use the formula for all of the three cases to find the required answer. We have to use the formula for all of the three cases to find the required answer. A combination is defined as a method or way of arranging elements from a set of elements such that the order of arrangement does not matter.

Complete step-by-step solution:
A) Since in this case there are no restrictions, the number of $4$ person committees which are possible from a group of $9$ people will be equal to the number of combinations of $4$ objects from a total of $9$ objects. Thus, the number of committees in this case is given by
${{n}_{1}}={}^{9}{{C}_{4}}$
We know that ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So we get
${{n}_{1}}=\dfrac{9!}{4!\left( 9-4 \right)!}$
$\Rightarrow {{n}_{1}}=\dfrac{9!}{4!5!}$
We can write $9!=9\times 8\times 7\times 6\times 5!$ and $4!=4\times 3\times 2\times 1$ in the above equation to get
${{n}_{1}}=\dfrac{9\times 8\times 7\times 6\times 5!}{4\times 3\times 2\times 1\times 5!}$
$\Rightarrow {{n}_{1}}=126$
Hence, the number of committees possible is equal to $126$.
B) In this case, two persons John and Barbara must necessarily be in the committee. So the number of seats is reduced from $4$ to $2$. So now we have to select $2$ persons out of the remaining $7$ persons to get
${{n}_{2}}={}^{7}{{C}_{2}}$
$\Rightarrow {{n}_{2}}=\dfrac{7!}{2!5!}$
Writing $7!=7\times 6\times 5!$, we get
${{n}_{2}}=\dfrac{7\times 6\times 5!}{2!5!}$
$\Rightarrow {{n}_{2}}=21$
Hence, in this case the number of possible committees is equal to $21$.
C) In this case we have to choose one of John or Barbara. So there are two ways of choosing the first person of the committee. The remaining $3$ persons are also to be selected from the remaining $8$ persons. Therefore, the number of committees in this case is given by
${{n}_{3}}=2\times {}^{8}{{C}_{3}}$
$\Rightarrow {{n}_{3}}=2\times \dfrac{8!}{3!5!}$
Writing $8!=8\times 7\times 6\times 5!$ and $3!=3\times 2\times 1$ we get
${{n}_{3}}=2\times \dfrac{8\times 7\times 6\times 5!}{3\times 2\times 1\times 5!}$
$\Rightarrow {{n}_{3}}=112$

Hence, the number of committees in this case is equal to $112$.

Note:
We should not consider the cases of the B and the C part to be mutually exclusive so that we can subtract the number of possible committees in part B from that of part A to obtain that of part C. This is because they both do not include the case when neither John nor Barbara is in the committee. We should not use the concept of permutation here because the permutation is a method of selecting a number of elements from a set of elements such that the order of selection matters.