Question

Six different sweets are to be distributed among 3 room mates $A_1,B_1,C_1$. The number of ways in which room mate $B_1$ gets 2 sweets is

• A. 240
• B. ${}^6{C}_3$
• C. ${}^6{C}_3\times 2^3$
• D. None of these

Answer 1

The correct option is A 240

Let's just give $B_{1}2$ sweets.This can be given in $\left(\begin{array}{c}6\\ 2\end{array}\right)$ ways.Now we have to distribute remaining 4 sweets in A and C.
This can be done in the following ways:$4\equiv (4,0)or4\equiv (1,3)or4\equiv (2,2)$
So the number of ways of this is respectively $\frac{4!}{4!}\times 2!,\frac{4!}{3!\times 1!}\times 2!,\frac{4!}{2!\times 2!}\times \frac{1}{2!}\times 2!,$
(we are multiplying by 2! because the sweets are to be distributed between 2 people A and C)
Hence the total number of ways =$\left(\begin{array}{c}6\\ 2\end{array}\right)\times (\frac{4!}{4!}\times 2!+\frac{4!}{3!\times 1!}\times 2!+\frac{4!}{2!\times 2!}\times \frac{1}{2!}\times 2!)=15\times (2+8+6)=240$