The question stem should make clear that the 5 marbles are DISTINCT:

2mist wrote:Q. In how many ways you can distribute 5 different marbles among 3 identical baskets such that each basket has at least one marble?

a. 10
b. 16
c. 25
d. 32
e. 64

Let the 5 marbles be A, B, C, D and E.

Case 1: 3 marbles together, 1 marble alone, 1 marble alone
From the 5 marbles, the number of ways to choose 3 marbles for the 1st basket = 5C3 = (5*4*3)/(3*2*1) = 10.
For each of these 10 cases, there is only 1 possible distribution.
For example, if ABC are together in 1 basket, the marbles must be divided as follows:
ABC-D-E.
Since there is only 1 possible distribution for each of the 10 combinations of 3, the total number of ways to distribute the marbles in Case 1 = 10.

Case 2: 1 marble alone, 2 marbles together, 2 marbles together
Number of options for the marble alone = 5. (Any of the 5 marbles.)
If A is the marble alone, here are all of the ways to distribute the marbles:
A-BC-DE
A-BD-CE
A-BE-CD
Total ways = 3.
Since there are 3 ways when A is alone, there will be 3 ways when B is alone, 3 ways when C is alone, 3 ways when D is alone, and 3 ways when E is alone.
Thus, the total number of ways to distribute the marbles in Case 2 = 3+3+3+3+3 = 15.

Total ways = 10+15 = 25.

The correct answer is C.