Question

The number of ways in which $8$ non-identical apples can be distributed among $3$ boys such that every boy should get atleast $1$ apple and atmost $4$ apples is a four digit number of the form $pqrs$ then $p.q.r$ equals.

Answer 1

No. of combinations of n dissimilar things taken r at a time is denoted by ${}^n{\text{C}}_r$.

${}^n{\text{C}}_r=\frac{n!}{r!(n-r)!}$

There are 3 cases for this:

$\left(a\right)$ Apples are distributed as $(4,3,1)$. This combination can be distributed in $3!$ ways.

No.of ways = $3!\times {}^8{C}_4\times {}^4{C}_3=6\times 70\times 4=1680$

$\left(b\right)$ Apples are distributed as $(4,2,2)$. This combination can be distributed in ${}^3{C}_2$ ways.

No.of ways = ${}^3{C}_2\times {}^8{C}_4\times {}^4{C}_2=3\times 70\times 6=1260$

$\left(c\right)$ Apples are distributed as $(3,3,2)$. This combination can be distributed in ${}^3{C}_2$ ways.

No.of ways = ${}^3{C}_2\times {}^8{C}_3\times {}^5{C}_3=3\times 56\times 10=1680$

So, the total number of possibilities = $1680+1260+1680=4620$.

Thus, if $4620$ is $pqrs$, then $p.q.r=4\times 6\times 2=48$.

$\therefore$ The correct answer is $48$.