Among three different boxes, 10 identical balls have to be distribute

Question

Among three different boxes, 10 identical balls have to be distributed. In how many ways can this be done such that every box has at least 2 balls?

(A) 15 
(B) 16 
(C) 64 
(D) 81 
(E) None of these

DashingNightmare2 · Accepted Answer

Let a,b,c are 3 boxes
since all contains atleast 2 balls => a+2 + b+ 2 + c+ 2 =10
=>a+b+c=4
number of ways= (n+r-1)C(r-1)
here n=4 ,r=3
Hence 6c2= 15 
IMO A.

TestPrepUnlimited · Answer

We need to distribute 10 balls among 3 boxes, and each box must have at least 2 balls.  Start by distributing 6 identical balls among 3 boxes.

Then we need to distribute 4 more balls among any of the 3 boxes. We can have 4 of the same box (3 cases), 1 + 1 + 2 (3 cases), 2 + 2 (3 cases), or 3 + 1 + 0 (6 cases). Thus 15 cases total.

Ans: A

GMATWhizTeam · Answer

Solution:

Since each box has to have at least 2 balls so lets us assume that we do so.

Now we have 4 balls that we have to keep in 3 boxes. Let us see the cases and get the number of ways.

Case 1:   4 balls in 1 box, 0 and 0 in other 2 boxes. For example:  4, 0, 0 or 0, 4, 0 or 0, 0, 4 . 
Number of ways =  \frac{3!}{2!} = 3

Case 2:   3 balls in 1 box, 1 in another and 0 in last box. For example:  3, 1, 0 or 3, 0, 1  and so on. 
Number of ways =  3! = 6

Case 3:   2 balls in 1 box, 1 in another box and 1 in last box. For example:  2, 1, 1 or 1, 2, 1  and so on. 
Number of ways =  \frac{3!}{2!} = 3

Case 4:   2 balls in 1 box, 2 in another box and 0 in last box. For example:  2, 2, 0 or 0, 2, 2  and so on. 
Number of ways =  \frac{3!}{2!} = 3

Total number of cases  = 3+6+3+3 = 15 .

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Among three different boxes, 10 identical balls have to be distribute

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Among three different boxes, 10 identical balls have to be distribute

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