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 $n\ge \sum _{i=1}^r{m}_i$.

Short answer

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

Step-by-step solution

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 arerole="math" localid="1648301319689" $=n-\sum _{i=1}^r{m}_i$

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 role="math" localid="1648301285511" $r$urns.

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