– Calculate the number of ways a departmental committee of five members can be selected, given that it must consist of at least one woman.
– Calculate the number of ways a departmental committee of five members can be selected, given that it must consist of at least one woman and one man.
The aim of this question is to find the number of ways for which a committee of a total of $5$ members should have at least $1$ woman. For the other part, we have to find a total number of ways for the committee to have one woman and one man.
In order to solve this problem the right way, we need to understand the concept of Permutation and Combination. A combination in mathematics is the arrangement of its given members irrespective of their order.
\[C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\]
$C\left(n,r\right)$ = number of combinations
$n$ = total number of objects
$r$ = selected object
A permutation in mathematics is the arrangement of its members in a definite order. Here, the order of the members matters and are arranged in a linear manner.
\[nP_r\\=\frac{n!}{\left(n-r\right)!}\]
$n$ = total number of objects
$r$ = selected object
$nP_r$ = permutation
It is an Ordered Combination. The difference between the two is in order. For example, the PIN of your mobile is $6215$, and if you enter $5216$, it won’t unlock as it is a different order (permutation).
Expert Answer
$(a)$ To find out the number of ways to select a committee of $5$ members with at least one woman, we will subtract the committees with only men from the total number of committees. Here, as the order of members does not matter, we will use a combination formula to solve this problem.
Total women = $7$
Total men = $9$
Total number of people= $7+9 =16$
$n=16$
The committee should be composed of $5$ members, $r=5$:
\[C\left(16,5\right)=\frac{16!}{5!\left(16-5\right)!}\]
\[C\left(16,5\right)=\frac{16!}{5!11!}\]
\[C\left(16,5\right)=4368\]
To select $5$ members from $9$ men:
$n= 9$
$r= 5$
\[C\left(9,5\right)=\frac{9!}{5!\left(9-5\right)!}\]
\[C\left(9,5\right)=\frac{9!}{5!11!}\]
\[C\left(9,5\right)=126\]
The total number of ways to select a committee of $5$ members with at least one woman is $=4368-126=4242$
$(b)$ To find out the number of ways to select the committee of $5$ members with at least one woman and one man, we will subtract the committees with only women and men from the total.
Committees with only women is given as:
$n= 7$
$r= 5$
\[C\left(7,5\right)=\frac{7!}{5!\left(7-5\right)!}\]
\[C\left(7,5\right)=\frac{7!}{5!2!}\]
\[C\left(7,5\right)=21\]
The number of ways to select the committee of $5$ members with at least one woman and at least one man = $4368 – 126 -21=4221$.
Numerical Results
The number of ways to select the committee of $5$ members with at least one woman is $4242$.
The number of ways to select the committee of $5$ members with at least one woman and at least one man is $4221$.
Example
A group of $3$ athletes is $P$, $Q$, $R$. In how many ways can a team of $2$ members be formed?
Using Combination formula:
$n=3$
$r=2$
\[C\left(3,2 \right)=\frac{3!}{2!\left(3-2\right)!}\]
\[C\left(3,2 \right)=3\]