Counting

Hi talaangoshtari,

This is an interesting 'concept' question; you can deal with it in a number of different ways, but if you don't see a 'formulaic' approach, then there is still a 'brute force' way to solve it. You'll have to take some notes and discover the built-in patterns though.

We're told that 10 duplicate balls will be put into 3 DISTINGUISHABLE bins and that each bin MUST have AT LEAST 1 ball in it. We're asked how different ways we can distribute those balls into those bins.

I'm going start with an example:
Let's split the balls into groupings of 1, 1 and 8

Since the bins are distinguishable, there are 3 ways to put these groups into the bins:
118
181
811

Now, what if our grouping did NOT contain a duplicate number.....
Let's split the balls into groupings of 1, 2 and 7

Here, we have 6 ways to put these groups into the bins:
127
172
217
271
712
721

THOSE patterns are what we need to figure out the TOTAL number of options. Now, we just have to figure out how many groupings there are....

118
127
136
145
226
235
244
334

There are 4 groupings with a duplicate number (3 options for each of those groups) and 4 groupings with no duplicate numbers (6 options for each of those groups).

4(3) + 4(6) = 12+24 = 36 different ways to distribute the balls.

Final Answer: D

GMAT assassins aren't born, they're made,
Rich

talaangoshtari wrote:In how many ways we can distribute 10 indistinguishable balls to 3 distinguishable bins that in each bin we have at least 1 ball?

A.25
B.15
C.40
D.36
E.18

I'd say that this question is a LITTLE too difficult/time-consuming for the GMAT.
That said, we can solve it using an approach that's similar to this question: https://www.beatthegmat.com/very-tricky- ... 25349.html

First, place 1 ball in each bin.
We've now taken care of the requirement that each bin has AT LEAST 1 ball.
So, the question now becomes, "In how many ways can we place the remaining 7 indistinguishable balls into 3 distinguishable bins?"

I'm not going to get into the logistics of the solution (since I feel it's out of scope), but using an approach that's similar to the one above at https://www.beatthegmat.com/very-tricky- ... 25349.html, the answer becomes 9C2 (which equals 36)

Cheers,
Brent