Question

There are 10 identical red balls, 10 identical blue balls, 10 identical green balls and 2 distinct boxes. How many ways to place all the 30 balls in the two boxes such that each box contains 15 balls?

Answer 1

Solution:

We are given that, two boxes are distinct.

Suppose r, y and b be the numbers of red, yellow and blue balls in the first box, respectively. Then we will find the number of cases satisfying r+y+b =15 and then subtract the numbers of cases where r > 10,y > 10 or b > 10.

Then by using stars and bars theorem,the number of ways to place n indistinguishable balls into k labelled urns = ^(n+k-1)C_(k-1) .

So the number of cases satisfying r + y+ b =15 is = (15+3-1)C(3-1) = 17C2

Using combination formula, 17C2 = 17!/(15!2!) = (17*16*15!)/(15!*2!) = (17*16)/2 = 17*8 = 136

Now for numbers of cases where r > 10.

If r = 11, then y+b = 4 implies that there are 5 such cases.                                                                If r = 12, then y+b = 3 implies that there are 4 such cases.                                                               If r = 13, then y+b = 2 implies that there are 3 such cases.                                                               If r = 14, then y+b = 1 implies that there are 2 such cases.                                                                If r = 15, then y+b = 0 implies that there are 1 such cases.

So number of cases when r > 10 is = 5+4+3+2+1 = 15. Similarly for y >10,                               number of cases = 5+4+3+2+1 = 15 and for b >10 number of cases = 5+4+3+2+1 = 15,                   So total cases = 15+15+15 = 45

Number of ways to place all 30 balls in two boxes such that each box contain 15 balls = 136 - 45 = 91