– 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 $r$:
$n=3$
$r=2$
\[C\left(3,2 \right)=\frac{3!}{2!\left(3-2\right)!}\]
\[C\left(3,2 \right)=3\]