Five balls of different colors are to be placed

[dddanny2006]

Five balls of different colors are to be placed in three boxes of different sizes. Each box can hold all five balls. In how many different ways can we we place the balls so that no box remains empty?

A. 150 B. 10 C. 60 D. 300 E. 375

We have 5 balls to be placed in 3 boxes and each box must have atleast one ball.

To get maximum possibilities-Lets split it this way

First box gets 5c3 ,the second one gets 2c1 and the third one gets 1c1 and the 3 boxes can be arranged in 3! ways

5c3*2c1*1c1*3! =120

Now lets proceed to the 2-2-1 possibility

5c2*3c2*2c1*3! =180

Where am I going wrong?I mean Im multiplying it by 3 factorial because the contents of the boxes can be rearranged in 3! ways.I may be wrong because Im multiplying each of those by 3! instead of just multiplying by 3.Can you please clear my confusion as to when do we multiply by the 3! and when just 3,Ive solved a lot of problems where we include the "!" sign and multiply it by 3!,so please clear my confusion.

Thanks.

[RonPurewal]

Hello,

Per the forum rules, please give the source of the problem. Thanks.

[dddanny2006]

Source is Veritas

Hello,

Per the forum rules, please give the source of the problem. Thanks.

[RonPurewal]

In each of these cases, you shouldn't multiply by 3!; you should just multiply by 3.

Think about what this number represents. It represents the number of ways you can "scramble" the boxes, so that different things are happening.

If you actually had something distinct happening in each box -- e.g., you have 6 balls, and you put 3 of them in one box, 2 in the next, and 1 in the last -- then you'd have 3! different possibilities.

Here, though, you don't.
In your 3-1-1 case, the two boxes getting 1 ball each are interchangeable, so, if you say there are 3! combinations, you're doubling the actual number.
Just realize that there's one "special" box -- the one that gets the 3 balls -- and there are exactly 3 possibilities for which box that can be.
Similarly, in the 2-2-1 case, the two boxes that get 2 balls each are also interchangeable. There's only one "special" box -- the one that gets the single ball -- so there are 3 arrangements of the boxes, depending on which box is the "special" one.

[RonPurewal]

By the way, this problem is much harder than anything I've ever seen on an actual GMAC problem. I hope that it appeared in some sort of challenge-problem bank, or something like that.

[dddanny2006]

Thanks for the explanation Ron.
I still get confused as to when we got to include the ! sign and when we got to just multiply by 3.
Ive solved problems where we just multiply including the factorial sign,but here we just multiply by 3.Cant all the 3 boxes be scrambled instead of just 1 box?

I can do about 60% of the problems relating to Combinatorics and Probability.My worst fear is that -my inability to solve the 700-800level problems in the same areas(Combinatorics and Probability) may stop me from getting that 700+ figure.To get a 700+,is it absolutely necessary to battle such combinatorics problem.

[jlucero]

Thanks for the explanation Ron.
I still get confused as to when we got to include the ! sign and when we got to just multiply by 3.
Ive solved problems where we just multiply including the factorial sign,but here we just multiply by 3.Cant all the 3 boxes be scrambled instead of just 1 box?

I can do about 60% of the problems relating to Combinatorics and Probability.My worst fear is that -my inability to solve the 700-800level problems in the same areas(Combinatorics and Probability) may stop me from getting that 700+ figure.To get a 700+,is it absolutely necessary to battle such combinatorics problem.

First, let me echo Ron's point about this being an extremely difficult question- one you'd be highly unlikely to see on an actual GMAT.

Second, you absolutely do not need to know how to answer questions like this in order to get a 700. In fact, you don't really need to answer any 700-800 level questions to get a 700. Your goal should be perfection on any 200-690 level questions and hope that you end your test on a high note. Getting a few 700+ level questions right has happened to most of our instructors, who have still scored in the 760+ range.

Finally, to answer your question about why you're not multiplying by 3!, it's because you're technically multiply by 3c2. When you say that one box will have 3 balls and two boxes will have 1 ball, there are 3!/2!1! ways of splitting this up. You can also think about this logically by saying there are 3 different boxes that could 3 balls, and the other two will just have 1. Your harder question is how to avoid doing this in the future, and unfortunately the answer is just more practice. Take solace in knowing that these are some of the hardest questions you'll see in your GMAT preparation. Because in order to differentiate between a 780 and a 790 GMAT scorer, the GMAT needs some REALLY hard questions. But unless your goal is to score 800, you can get questions wrong and still do great on this test.