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.

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  $u\ge 0, v\ge 0, w\ge 0 then$

$\therefore$ x + y + z =21

$\Rightarrow$ u + 5 + v + 5 + w + 5 = 21

$\Rightarrow$ u + v + w = 6 

$\therefore$Total number of solutions =  $C3-16+3-1=C28=28$