smanstar wrote:In how many ways can you distribute 30 identical balls in 7 boxes (B1, B2..., B7)
Such that :
No 2 boxes have same no of balls and at-least 1 ball is in each box ?

1+2+3+4+5+6+7 = 28.
This represents the MINIMUM number of balls needed to satisfy the conditions that no box is empty and that each of the 7 boxes receives a different number of balls.
Since there are a total of 30 balls, 2 balls remain to be distributed.

Case 1: The remaining 2 balls are placed in the box with 7 balls
1+2+3+4+5+6+(7+2) = 1+2+3+4+5+6+9 = 30.
Thus, any arrangement of the seven integers 1, 2, 3, 4, 5, 6 and 9 represents a valid way to distribute the 30 balls.
Number of ways to arrange these seven integers = 7! = 5040.

Case 2: The remaining 2 balls are placed in the box with 6 balls
1+2+3+4+5+(6+2)+7 = 1+2+3+4+5+7+8 = 30.
Thus, any arrangement of the seven integers 1, 2, 3, 4, 5, 7 and 8 represents a valid way to distribute the 30 balls.
Number of ways to arrange these seven integers = 7! = 5040.

There is no other way to distribute the remaining 2 balls so that each box contains a different number of balls.
For example, if the 2 remaining balls are placed in the box with 5 balls, we get:
1+2+3+4+(5+2)+6+7 = 1+2+3+4+6+7+7.
Here, two boxes have 7 balls each, so this distribution is not valid.

Total valid distributions = 5040+5040 = 10,080.

GMATGuruNY wrote:

smanstar wrote:In how many ways can you distribute 30 identical balls in 7 boxes (B1, B2..., B7)
Such that :
No 2 boxes have same no of balls and at-least 1 ball is in each box ?

1+2+3+4+5+6+7 = 28.
This represents the MINIMUM number of balls needed to satisfy the conditions that no box is empty and that each of the 7 boxes receives a different number of balls.
Since there are a total of 30 balls, 2 balls remain to be distributed.

Case 1: The remaining 2 balls are placed in the box with 7 balls
1+2+3+4+5+6+(7+2) = 1+2+3+4+5+6+9 = 30.
Thus, any arrangement of the seven integers 1, 2, 3, 4, 5, 6 and 9 represents a valid way to distribute the 30 balls.
Number of ways to arrange these seven integers = 7! = 5040.

Case 2: The remaining 2 balls are placed in the box with 6 balls
1+2+3+4+5+(6+2)+7 = 1+2+3+4+5+7+8 = 30.
Thus, any arrangement of the seven integers 1, 2, 3, 4, 5, 7 and 8 represents a valid way to distribute the 30 balls.
Number of ways to arrange these seven integers = 7! = 5040.

There is no other way to distribute the remaining 2 balls so that each box contains a different number of balls.
For example, if the 2 remaining balls are placed in the box with 5 balls, we get:
1+2+3+4+(5+2)+6+7 = 1+2+3+4+6+7+7.
Here, two boxes have 7 balls each, so this distribution is not valid.

Total valid distributions = 5040+5040 = 10,080.

Hi Mitch

Can we not distribute the 2 remaining balls with the 6th and 7th ball getting 1 ball each ?

smanstar,

even if we distribute the 2 remaining balls with the 6th and 7th ball getting 1 ball each,
1+2+3+4+5+6+7 = 28.
this will become
1+2+3+4+5+7+8 = 28.

which is already been accounted in case 2.