**– There are $25$ members in a club.**

**– In how many ways can $4$ members be chosen to serve in an executive committee?**

**– In how many ways can a president, vice president, secretary, and treasurer of the club be chosen so that each person can only hold a single office at a time?**

The aim of this question is to find the **number of ways for which an executive committee can be served by $4$ members**.

For the other part, we have to find a **number of ways to choose a president, vice president, etc without giving the same position to $2$ members**

In order to correctly solve this problem, 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 is arranged in a **linear manner.** It is also called an **Ordered Combination**, and 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 ordering **(permutation).**

\[nP_r\\=\frac{n!}{\left(n-r\right)!}\]

**$n$ = Total number of objects**

**$r$ = Selected object**

**$nP_r$ = Permutation**

## Expert Answer

**$(a)$ Find the number of ways for which an executive committee can be served by $4$ members. **Here, as the order of members does not matter, we will use **combination formula.**

$n=25$

The committee should be of $4$ members, $r=4$

\[C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\]

Putting values of $n$ and $r$ here, we get:

\[C\left(25,4\right)=\frac{25!}{4!\left(25-4\right)!}\]

\[C\left(25,4\right)=\frac{25!}{4!21!}\]

\[C\left(25,4\right)=12,650\]

**The number of ways to select the committee of $4$ members** $=12,650$

**$(b)$ To find out the number of ways to select the club members for a president, vice president, secretary, and treasurer of the club,** the order of members is significant, so we will use the definition of **permutation**.

Total number of club members $=n=25$

Designated positions for which members are to be selected $=r=4$

\[P\left(n,r\right)=\frac{n!}{\left(n-r\right)!}\]

Putting values of $n$ and $r$:

\[P\left(25,4\right)=\frac{25!}{\left(25-4\right)!}\]

\[P\left(25,4\right)=\frac{25!}{21!}\]

\[P\left(25,5\right)=\frac{25 \times 24 \times 23 \times 22 \times 21!}{21!}\]

\[P\left(25,5\right)=25 \times 24 \times 23 \times 22\]

\[P\left(25,5\right)=303,600\]

**The number of ways to select the club members for a president, vice president, secretary, and treasurer of the club** $=303,600$.

## Numerical Results

The **number** of **ways** to choose $4$ **members** of the club to serve on an **executive committee** is $12,650$

The number of ways to select the club members for a **president, vice president, secretary,** and **treasurer** so that no person can hold more than one office is $303,600$.

## Example

A **group** of $3$ athletes is $P$, $Q$, $R$. In how many ways can a **team** of $2$ members be formed?

Here, as the **order** of **members** is not important, we will use the **Combination formula.**

\[C\left(n,r\right)=\frac{n!}{r!\left(n-r\right)!}\]

Putting values of $n$ and $r$:

$n=3$

$r=2$

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

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