The answer to this week’s question:

**The correct answer is A.**

Video Explanation:

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Hi Chris, I’m confused as to why it’s C3 and not C6, if there are 6 shirts, why are we not using 6C6?

That’s a great question – and this is a tricky problem.

Essentially, we are choosing three green shirts from a total of six. Let’s say those kids who get a green shirt are represented by an ‘X'; those who get a red shirt by an ‘O’.

X X X O O O

Notice that if the first three kids receive a green shirt, as is noted above, the kids in the 4th, 5th, and 6th position have to get red shirts, basically by default, i.e., they did not receive the green shirts. Above is just one scenario. Let’s say we have:

O O X O X X

Now the 3rd, 5th, and 6th kids are receiving green shirts. This is one unique circumstance, just like the first example is. We don’t have to see how many different cases exist in which we can distribute the red shirts, once the 3rd, 5th, and 6th kids have the green shirts; the only case in which red shirts are distributed is to the 1st, 2nd, and 4th student.

So essentially, we can discount handing out the three red shirts and just focus on the green shirts. This gives rise to 6C3.

Hope that makes sense :)!

Thank you Chris, that was very helpful as usual!

Hello,

How would we solve the quesiton if it read as follows-

In how many different ways can 4 identical green shirts and 2 identical red shirts be distributed among 6 children such that each child receives a shirt|?

thanks

sandesh.

In this case we would just have 6C4. Once you’ve determined who has received the green shirts, the red shirts automatically have to go to the remaining two. So the answer is 6C4 = 15.

Hope that helps :).

Hi Chris,

Thanks. It could also be viewed as selecting 2 kids in 6C2 ways to determine how many of them receive the green shirts, which amounts to 15 ways.

Exactly, the same result, and essentially the same problem :).