Question

In how many ways $12$ apples can be distributed among $4$ children such that every child gets at least $2$ apples?

Answer 1

Since, it is given that each of them should be given atleast $2$ apples.

Thus, firstly,we give $2$ to each of them (Assuming all apples are identical).

Now, these $4$ children have $2$ apples each.

Remaining $4$ apples ($12$ initial$–8$ distributed) can be distributed to these with the condition that anyone of these can receive any number of apples ranging from $0$ to $4$, as they will always have atleast $2$ with them as asked in the questions.

Let us now see the possible groups -

$1.4+0+0+0$

Number of ways $=\frac{4!}{3!}=4$

$2.3+1+0+0$

Number of ways $=\frac{4!}{2!}=12$

$3.2+2+0+0$

Number of ways $=\frac{4!}{2!2!}=6$

$4.2+1+1+0$

Number of ways $=\frac{4!}{2!}=12$

$5.1+1+1+1$

Number of ways $=\frac{4!}{4!}=1$

So, the total no. of possible ways $=4+12+6+12+1=35$ ways.