Q. 1.8 - A First Course in Probability

Question

Prove that: 

$\left(\begin{array}{c}n+m\\ r\end{array}\right)=\left(\begin{array}{c}n\\ 0\end{array}\right)\left(\begin{array}{c}m\\ r\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)\left(\begin{array}{c}m\\ r-1\end{array}\right)+..........+\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ 0\end{array}\right)$

Hint: Consider a group of $n$ men and $m$ women. How many groups of size $r$ are possible?

Step-by-Step Solution

Answer

The possible number of groups are $\left(\begin{array}{c}n+m\\ r\end{array}\right)=\left(\begin{array}{c}n\\ 0\end{array}\right)\left(\begin{array}{c}m\\ r\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)\left(\begin{array}{c}m\\ r-1\end{array}\right)+..........+\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ 0\end{array}\right)$.

Step 1. Given information.

Here, it is given that

No. of men $=n$

No. of women $=m$

Total no. of members in group $=n+m$

A sub group of $r$ members is to be formed.

Step 2. Find the possible combinations.

The possible combinations of a group of $r$ members are -

Case 1: Selecting $r$ women from the group of women and $0$ men from the group of men $={}^{n}C_0\times {}^{m}C_r$

Case 1: Selecting $r-1$ women from the group of women and $1$ men from the group of men $={}^{n}C_1\times {}^{m}C_{r-1}$

Case 1: Selecting $r-2$ women from the group of women and $2$ men from the group of men $={}^{n}C_2\times {}^{m}C_{r-2}$

.....

.....

Case 1: Selecting $1$ women from the group of women and $r-1$ men from the group of men $={}^{n}C_{r-1}\times {}^{m}C_1$

Case 1: Selecting $0$ women from the group of women and $r$ men from the group of men $={}^{n}C_r\times {}^{m}C_0$

Step 3. Find the required possibilities.

The possible number of ways of selecting $r$ members from the group of $n+m$ members will be

$\left(\begin{array}{c}n+m\\ r\end{array}\right)={}^{n}C_0\times {}^{m}C_r+{}^{n}C_1\times {}^{m}C_{r-1}+{}^{n}C_2\times {}^{m}C_{r-1}+..........+{}^{n}C_{r-1}\times {}^{m}C_1+{}^{n}C_r\times {}^{m}C_0$

Hence it is proved that

$\left(\begin{array}{c}n+m\\ r\end{array}\right)=\left(\begin{array}{c}n\\ 0\end{array}\right)\left(\begin{array}{c}m\\ r\end{array}\right)+\left(\begin{array}{c}n\\ 1\end{array}\right)\left(\begin{array}{c}m\\ r-1\end{array}\right)+..........+\left(\begin{array}{c}n\\ r\end{array}\right)\left(\begin{array}{c}m\\ 0\end{array}\right)$