Question

In how many ways can $7$ identical balls be placed into four boxes $P,Q,R,S$ such that the two boxes P and Q have at least one ball each?

• A. $84$
• B. $70$
• C. $120$
• D. $56$

Answer 1

The correct option is D $56$

If there are $k$ indistinguishable balls that one can put into n distinguishable boxes, then there are ${}^{n+k-1}{C}_k$ ways to place them.

As $P$ and $Q$ boxes should have at least one ball in each, so we put one in each of them at then we are left with $5$ balls.

So now $k=5$ and $n=4$ so the number of ways are ${}^{4+5-1}{C}_5=\frac{8!}{5!\times 3!}=8\times 7=56$