**– 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:**

$n=3$

$r=2$

\[C\left(3,2 \right)=\frac{3!}{2!\left(3-2\right)!}\]

\[C\left(3,2 \right)=3\]