Mission2012 wrote:In how many ways can 5 different marbles be distributed in 4 identical pockets?
(A) 24
(B) 51
(C) 120
(D) 625
(E) 1024
Since the pockets are IDENTICAL, where the marbles are placed is irrelevant.
All that matters is how the marbles are DIVIDED.
Case 1: All 5 marbles in one pocket
Number of ways to choose 5 marbles from 5 options = 5C5 = 1.
Case 2: 4 marbles in one pocket, 1 marble alone
Number of ways to choose 4 marbles from 5 options = 5C4 = 5.
The remaining marble must be alone.
Total options = 5.
Case 3: 3 marbles in a one pocket, 2 marbles in another
Number of ways to choose 3 marbles from 5 options = 5C3 = 10.
The remaining 2 marbles must be together.
Total options = 10.
Case 4: 3 marbles in one pocket, 2 marbles each alone
Number of ways to choose 3 marbles from 5 options = 5C3 = 10.
The remaining 2 marbles must be in separate pockets.
Total options = 10.
Case 5: 2 marbles in one pocket, 2 marbles in another pocket, 1 marble alone
Number of ways to choose 2 marbles from 5 options = 5C2 = 10.
Number of ways to choose 2 marbles from the remaining 3 options = 3C2 = 3.
The remaining marble must be alone.
To combine the options above, we multiply:
10*3.
Here, we must be careful.
Because the pockets are IDENTICAL, where the two pairs are placed is irrelevant.
Thus, the ORDER of the pairs doesn't matter: AB-CD is the same way of dividing the marbles as CD-AB.
Since the order of the pairs doesn't matter, the result above must be divided by the number of ways the 2 pairs can be ARRANGED (2!):
Total options = (10*3)/(2*1) = 15.
Case 6: 2 marbles in one pocket, 3 marbles each alone
Number of ways to choose 2 marbles from 5 options = 5C2 = 10.
The remaining 3 marbles must each be alone.
Total options = 10.
Total ways = 1+5+10+10+15+10 = 51.
The correct answer is
B.
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