Question

Number of ways in which $5$ different toys can be distributed among $5$ children if exactly one child do not get any toy is:

• A. $1100$
• B. $1200$
• C. $1300$
• D. $240$

Answer 1

Answer: (B)

$1200$

There are $5$ ways to choose the child who did not get any toy. There are

${}^5{C}_2$ ways to choose $2$ toys as only $1$ child can get max $2$ toys.

There are $5-1=4$ ways to choose the luckiest child, who will get $2$ toys.

There are $(5-2)!=3!$ ways to distribute the rest $3$ toys to the remaining $3$ children.

Hence, total no. of ways $=5\times 4\times 3!{\times }^5{C}_2=120\times 10=1200$