parveen110 wrote:In how many ways can 73 identical balls be put into three boxes - B1, B2 and B3 - such
that B1 contains more balls than B2 and B2 contains more balls than B3?

(a) 888
(b) 444
(c) 222
(d) 666
(e) 999

OA:444

This problem is WAY beyond the scope of the GMAT.
That said, here is my solution:

Use the SEPARATOR method.
For an explanation of this method, check here:
https://www.beatthegmat.com/lets-have-a- ... 69973.html

Here, 73 identical balls are to be distributed among 3 bags.
Thus, we need to count the number of ways to arrange 73 identical balls and 2 identical separators (for a total of 75 elements):
75!/(73!2!) = 75*37 = 2775.

Each of the distributions above must be in DESCENDING ORDER.
Thus, we must subtract from the total the BAD CASES that are NOT in descending order.

Bad Case 1: 2 of the bags receive the SAME NUMBER of marbles
WRITE IT OUT and LOOK FOR A PATTRN.
The following distributions are not viable:
73-0-0, 0-73-0, 0-0-73 --> 3 cases
71-1-1, 1-71-1, 1-1-71 --> 3 cases
This pattern will continue until:
1-36-36, 36-1-36, 36-36-1-> 3 cases.
Notice that the values in red constitute the odd integers between 1 and 73, inclusive, for a total of 37 integers.
Implication:
For each of these 37 integers, there will be 3 bad cases:
37*3 = 111.
Subtracting these 111 bad cases from the total, we get:
2775 - 111 = 2664.

Bad Case 2: 3 distinct groupings NOT in descending order
Test an easy example.
If the 73 balls are divided into one group of 60, one group of 10, and one group of 3, the groupings can be arranged as follows:
B1 = 60, B2 = 10, B3 = 3
B1 = 60, B2 = 3, B3 = 10
B1 = 10, B2 = 60, B3 = 3
B1 = 10, B2 = 3, B3 = 60
B1 = 3, B2 = 60, B3 = 10
B1 = 3, B2 = 10, B3 = 60.
Only the distribution in red is in descending order..
Implication:
Of the remaining 2664 ways, 1 of every 6 will be viable.
Dividing 2664 by 6, we get:
2664/6 = 444.

The correct answer is D