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

Question

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

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

CrackverbalGMAT · Accepted Answer

First, Let us give each box 2 balls. 6 are distributed, and we now have 4 balls left.

The different ways of getting a sum of 4

4,0,0 : This can be arranged in  \frac{3!}{2!} = 3  ways (400, 040 and 004)

1,1,2 : This can be arranged in  \frac{3!}{2!} = 3  ways (112,121 and 211)

2,2,0 : This can be arranged in  \frac{3!}{2!} = 3  ways (220,202 and 022)

1,3,0 : This can be arranged in  3! = 6  ways (130,103, 301, 310, 013 and 031)

Therefore the total number of ways = 3 + 3 + 3 + 6 = 15

Option A

Arun Kumar

aritrar4 · Answer

We need to distribute at least 2 balls to each of the baskets B1, B2 and B3. Since the balls are identical, we can only distribute them in 1 way.

We now have 4 identical balls remaining, since 6 are distributed already.

So we need to divide 4 identical balls in 3 different baskets. If we use the "partition bar" method and add 2 bars between 4 balls we can find the list of all such arrangements. OO|O|O

So we have 6 total objects (4balls and 2 sticks to divide them). Therefore total arrangements = 6!/4!2! = 6*5/2 = 15 arrangements

IMO option A.

gmatcafe · Answer

In 15 ways:

6 2 2
5 3 2
5 2 3
4 3 3
4 4 2
4 2 4
3 2 5
3 5 2
3 4 3
3 3 4
2 2 6
2 6 2
2 5 3
2 3 5
2 4 4

Answer is A.

Akshaynandurkar · Answer

Lets call 3 boxes as A,B and C
A, B and C will get 2 balls each out of 10 balls . so we are left with 4 balls. 
These 4 balls can be distributed in following ways
1)4,0,0 ----3!/2! = 3 ways
2)1,1,2 ----3!/2! = 3 ways
3)2,2,0 ----3!/2! = 3 ways
4)3,1,0 ----3! = 6 ways
Total ways = 15 ways

PSKhore · Answer

Since all the balls are identical, they can be distributed in one way, so 6 balls are gone.
To calculate further, when distributing 'n' identical objects into 'r' distinct boxes, where each box can contain zero or more objects, the number of ways to distribute them is given by the combination formula (n+r-1)C(r-1)

(4+3-1)C(3-1) = 6C2 = 15

svsivakrishna · Answer

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

Since each box should contain atleast 2 balls, min 6 balls are already distributed. Therefore, the question can be re-written as to distribute 4 balls into 3 boxes with each box containing the ball count of 0 to 4.

Therefore,
it can be written ( let each ball be denoted "b" and to be divided into 3 boxes, therefore we have 2 separators say "s")- one possible combination is b s bb s b (1st box has 1 ball, 2nd has 2 and thrid has 1).

Number of possible combinations for arrangement of 4 b and 2 s is 6!/( 4!*2!) = 15

Option A

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

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

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