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Question

Fifteen identical balls have to be put in five different boxes. Each box can contain any number of balls. The total number of ways of putting the balls into the boxes so that each box contains at least two balls is equal to

A
9C5
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B
10C5
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C
6C5
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D
10C6
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Solution

The correct option is A 9C5
Concept: Total number of non-negative integral solution of x1+x2+......+xr=nisn+r1Cr1
Also, n identical things can be distributed in r groups in n+r1Cr1 ways

Let the balls put in the box are x1,x2,x3,x4 and x5. We have,
x1+x2+x3+x4+x5=15,xi2
(x12)+(x22)+(x32)+(x42)+(x52)=5
y1+y2+y3+y4+y5=5,yi=xi20
The total number of ways is equal to number of non-negative integral solutions of the last equation, which is equal to 5+51C5=9C5.

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