How many ways are there to choose four members of the club to serve on an executive committee?

– 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.

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

$C(n,r)$ = 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).

$$\begin{gathered} n{P}_r \\ =\frac{n!}{(n-r)!} \end{gathered}$$

$n$ = Total number of objects

$r$ = Selected object

$n{P}_r$ = 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, $r=4$

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

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

$$C(25,4)=\frac{25!}{4!(25-4)!}$$

$$C(25,4)=\frac{25!}{4!21!}$$

$$C(25,4)=12,650$$

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 $=r=4$

$$P(n,r)=\frac{n!}{(n-r)!}$$

Putting values of $n$ and $r$:

$$P(25,4)=\frac{25!}{(25-4)!}$$

$$P(25,4)=\frac{25!}{21!}$$

$$P(25,5)=\frac{25\times 24\times 23\times 22\times 21!}{21!}$$

$$P(25,5)=25\times 24\times 23\times 22$$

$$P(25,5)=303,600$$

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.

$$C(n,r)=\frac{n!}{r!(n-r)!}$$

Putting values of $n$ and $r$:

$n=3$

$r=2$

$$C(3,2)=\frac{3!}{2!(3-2)!}$$

$$C(3,2)=3$$

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