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Q51 - no. of ways to distribute 5 freshmen to 4 rooms such that none are empty.
Why is the answer (5C2)4! ?
Should it not be (5C2)(3C1)(2C1)4! ?


In Q28 which is similar - no. of ways to distribute 5 diff balls to 3 diff boxes such that none is empty, the method is
(5C2)(3C2)3! + (5C3)(2C1)3!

By the same logic, Q51 should be as mentioned above.
Plz respond quickly. Thx.

The correct answer is 5C2 * 4!.

Let us run through the logic of your answer - which is (5C2)(3C1)(2C1)4! ?.

The first 5C2 is to get a pair of freshmen to be roommates. Having done that, the action of multiplying that with 3C1 implies that you are trying to see which of the remaining 3 freshmen will be accomodated in ROOM NO. 2. The same logic goes for 2C1 for ROOM NO.3. And who ever is left out, occupies ROOM No.4.

If you took this approach, then where is the need to multiply all of these by 4!. You only need to multiply (5C2)(3C1)(2C1) by 4 and not 4!. Multiplying by 4 is to decide which of the 4 rooms will accomodate the two students who share a room.

Therefore, the final answer will be (5C2)(3C1)(2C1) * 4 = 5C2 * 3 * 2 * 4 = 5C2 * 4!.

In the way, you have approached you have willy nilly rearranged twice three of the freshmen who are provided single accomodations.

Hope this explanation helps.