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  4. In how many ways Ram can distribute 40 apples in his six children named A, B, C, D, E and F such that A gets two more than B, C gets 3 more than F and D gets five less than E and every one must have atleast one fruit

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Q. In how many ways Ram can distribute $40$ apples in his six children named $A, B, C, D, E$ and $F$ such that $A$ gets two more than $B, C$ gets 3 more than $F$ and $D$ gets five less than $E$ and every one must have atleast one fruit

Permutations and Combinations

Solution:

Let $n(B)=x n(A)=x+2$
$n(F)= y \,\,\,n(C)= y +3$
$n (D)= z \,\,\,n(E)= z +5$
$x+y+z+x+2+y+3+z+5=40$
$x+y+z=15$ but $x \geq 1, y \geq 1$ and $z \geq 1 .$
So total ways of distribution is
${ }^{14} C_{2}=91$