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

Question

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 (A) 101 (B) 91 (C) 96 (D) 136

Tardigrade Admin · Accepted Answer

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 ≥ 1, y ≥ 1 and z ≥ 1 . So total ways of distribution is 14 C2=91

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

101

91

96

136

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
$^{14} C_{2} = 91$

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

101

91

96

136

Let n(B)=xn(A)=x+2 n(F)=yn(C)=y+3 n(D)=zn(E)=z+5 x+y+z+x+2+y+3+z+5=40 x+y+z=15 but x≥1,y≥1 and z≥1. So total ways of distribution is 14C2​=91

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

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
$^{14} C_{2} = 91$