Suppose we have six marbles: 3 blue marbles, 2 red marbles,

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by Mike@Magoosh » Mon Feb 02, 2015 4:43 pm

Suppose we have six marbles: 3 blue marbles, 2 red marbles, and one green marble. Suppose we are going to put them into three cups: a black cup, a white cup, and a purple cup. The only restriction is that the two red marbles can't be in the same cup. We could put as many as five (all except one of the reds) in any cup. We could leave one cup empty, or put some in each of the three cups. All combinations are allowed that don't involve the two red marbles in the same cup. How many combinations are possible?
(A) 90
(B) 180
(C) 360
(D) 540
(E) 720

The GMAT loves tricky counting problems such as this. For a discussion of how to approach such problems, with the OA & OE to this particular problem, see:
https://magoosh.com/gmat/2015/counting- ... -the-gmat/

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Source: Beat The GMAT — Problem Solving | Mon Feb 02, 2015 4:43 pm

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by GMATGuruNY » Thu Feb 12, 2015 11:56 am

3 blue marbles:
Use the SEPARATOR METHOD -- a great way to count the number of ways n identical objects can be distributed among r distinct containers.
Here, 3 identical blue marbles are to be distributed among -- at most -- 3 distinct cups.
Thus, we need 3 marbles and 2 separators:
O|O|O

Any arrangement of the elements above represents one way to distribute the 3 identical marbles among 3 cups B, W and P.
O|O|O = B gets 1 marble, W gets 1 marble, P gets 1 marble.
OO||O = B gets 2 marbles, W gets 0 marbles, P gets 1 marble.
OOO|| = B gets all 3 marbles.
And so on.

To count all of the possible distributions, we simply need to count the number of ways to arrange the 5 elements above (the 3 identical marbles and the 2 identical separators).
The number of ways to arrange 5 elements = 5!.
But when an arrangement includes identical elements, we must divide by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change, reducing the total number of unique arrangements.
Here, we must divide by the number of ways to arrange the 3 identical marbles (3!) and the number of ways to arrange the 2 identical separators (2!):
5!/(3!2!) = 10.

2 red marbles:
Since the 2 red marbles cannot be in the same cup, we must choose 2 cups to contain 1 red marble each.
From 3 cups, the number of ways to choose 2 = 3C2 = (3*2)/(2*1) = 3.

Green marble:
Number of options for the 1 green marble = 3. (Any of the 3 cups.)

To combine the options for each colored marble, we multiply:
10*3*3 = 90.

The correct answer is A.

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