Question

Find the number of ways in which 21 balls can be distributed among 3 persons such that each person does not receive less than 5 balls.

• A. 28
• B. 14
• C. 21
• D. 7

Answer 1

The correct option is A 28

Let x, y, z be the number of balls received by the three persons, then
$x\ge 5,y\ge 5,z\ge 5$ and x + y + z = 21
Let x = u + 5 , y = v + 5 , z = w + 5
Let $u\ge 0,v\ge 0,w\ge 0$, then
$\therefore$ x + y + z = 21
$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21
$\Rightarrow$ u + v + w = 6
$\therefore$ Total number of solutions ${=}^{6+3-1}{C}_{3-1}{=}^8{C}_2=28$