Q. 1.20 - A First Course in Probability

Question

In how many ways can $n$ identical balls be distributed into $r$ urns so that the $ith$ urn contains at least $m_i$ balls, for each $i = 1, . . . , r$? Assume that $n\ge \sum _{i=1}^r{m}_i$.

Step-by-Step Solution

Answer

The required number of ways are = $n-\sum _{r-1}m+r-1$.

Step 1. Given information.

It is given that, 

No. of identical balls = $n$

No. of urns = $r$

The $ith$ urn contains at least $m_i$balls, where $i=1,2,3,......,r$.

Step 2. Find the required no. of ways.

Firstly, we distribute $m_i$ balls in the $ith$urn, where $i=1,2,3,......,r$.

So, the no. of balls that have been distributed $=\sum _{i=1}^r{m}_i$

The remaining balls are  $=n-\sum _{i=1}^r{m}_i$

By proposition 6.2, there are $\left(\begin{array}{c}n+r-1\\ r-1\end{array}\right)$

The distinct non-negative integer valued vectors $\left(x_1,x_2,.....,x_r\right)$ satisfying $x_1+x_2+.....+x_r=n$

Similarly in this case we have to distribute $n-\sum _{i=1}^r{m}_i$ balls in $r$ urns.

Therefore, it can be done in $n-\sum _{r-1}m+r-1$ ways.