The number of ways in which 21 identical apples can be distributed among three children such that each child gets at least 2 apples, is
- 130
- 136
- 142
- 406
Answer
(2) 136
Solution
Give 2 apples to each child first ⇒ 21-6=15 apples left
x1+x2+x3=15, xi 0
∵ non-negative integer solutions of x1+ +xr=n number n+r-1Cr-1
17C2= 17 162=136
Answer: (2)