Question

The number of ways in which $8$ distinguishable apples can be distributed among $3$ boys such that every boy should get atleast $1$ apple and atmost $4$ apples is $K$.${}^7{P}_3$ where $K$ has the value equal to ?

• A. $14$
• B. $66$
• C. $44$
• D. $22$

Answer 1

The correct option is D $22$

There are $3$ ways of distributing $8$ apples among boys, they are

$\text{1) 4 3 1}$

$\text{2) 3 3 2}$

$\text{3) 4 2 2}$

so number of ways$=3!{(}^8{C}_4\left){(}^4{C}_3\right){(}^1{C}_1)+\frac{3!}{2}{(}^8{C}_3){(}^5{C}_3\left)\right(2_2^C)+\frac{3!}{2}{(}^8{C}_4){(}^4{C}_2\left){(}^2{C}_2\right)=4620$

Now, $K{\times }^7{P}_3=4620⟹K=\frac{4620\times 4!}{7!}=22$