91Ó°ÊÓ

Total no. of persons $=n$

No. of persons to be chosen for committee $=k$

So, the number of members for the committee can be chosen in localid="1647947695401" ${}^{n}C_k$ways =$\left(\begin{array}{c}n\\ k\end{array}\right)$ways

Out of $k$members, one can be selected as chairperson in ${}^{k}C_1$ways = $k$ways

Therefore, total no. of ways are$\left(\begin{array}{c}n\\ k\end{array}\right)k$.

Out of $k$members, $1$member will be the chair person. So, the non-chair members will be $k-1$.

So, non-chair members can be selected out of $n$members in role="math" localid="1647948253033" ${}^{n}C_{k-1}$ways $=\left(\begin{array}{c}n\\ k-1\end{array}\right)$ways.

The left number of persons will be role="math" localid="1647948086924" $n-(k-1)=n-k+1$

Out of the left members, one can be selected as chairperson in ${}^{n-k+1}C_1$ways $=n-k+1$ways.

Therefore, total no. of ways are$\left(\begin{array}{c}n\\ k-1\end{array}\right)\left(n-k+1\right)$.

Out of $n$persons, one can be selected as chairperson in role="math" localid="1647948578979" ${}^{n}C_1$ways $=n$ways.

The remaining no. of people will be $=n-1$

The number of non-chair members of the committee will be $=k-1$

So, non-chair members can be selected out of $n-1$members in ${}^{n-1}C_{k-1}$ways = $\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$ways

Therefore, total no. of ways are$n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$.

In all the above parts, (a), (b) and (c), we have found the ways in which we could select $k$committee members out of the group of $n$persons and one chair person of the committee.

Therefore, the total number of ways will be equal.

Hence, it is proved thatrole="math" localid="1647950480646" $k\left(\begin{array}{c}n\\ k\end{array}\right)=\left(n-k+1\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)=n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$

As per the factorial definition, $\left(\begin{array}{c}m\\ r\end{array}\right)=\frac{m!}{r!(m-r)!}$.

Therefore,

$k\left(\begin{array}{c}n\\ k\end{array}\right)=k×\frac{n!}{k!(n-k)!}=k×\frac{n!}{k×(k-1)!(n-k)!}=\frac{n!}{(k-1)!(n-k)!}$................ (1)

$\left(n-k+1\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)=\left(n-k+1\right)×\frac{n!}{(k-1)!(n-k+1)!}=\left(n-k+1\right)×\frac{n!}{(k-1)!(n-k+1)×(n-k)!}=\frac{n!}{(k-1)!(n-k)!}$............ (2)

$n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)=n×\frac{(n-1)!}{(k-1)!(n-1-k+1)!}=\frac{n×(n-1)!}{(k-1)!(n-k)!}=\frac{n!}{(k-1)!(n-k)!}$............... (3)

As (1) = (2) = (3)

Therefore, it is proved that$k\left(\begin{array}{c}n\\ k\end{array}\right)=\left(n-k+1\right)\left(\begin{array}{c}n\\ k-1\end{array}\right)=n\left(\begin{array}{c}n-1\\ k-1\end{array}\right)$