In how many different ways can 3 identical green shirts and 3 identical red shirts be distributed among 6 children such that each child receives a shirt?

##### Title

Children's shirts

##### Your Result

Correct

##### Difficulty

Very Hard

##### Your Pace

0:01

##### Others' Pace

1:23

## Video Explanation

## Text Explanation

Think about the problem this way: I have 3 identical green shirts and 3 identical red shirts to distribute to six children. If I simply designate the three who will get green shirts, that determines everything, because once I know which three get the green shirts, I automatically know the other three get the red shirts. So, the number of possible outcomes is just the number of ways we could choose a combination of 3 from the set of six. That's the combination number nCr:

6C3 = 20.

**Frequently Asked Questions**:

**FAQ: Can I use the MISSISSIPPI Rule?**

**A: **Yes, you can!

We have 6 shirts, 3 are red and 3 are green. We can first treat this problem as if each shirt were unique. If that were the case, we'd use the Fundamental Counting Principle, and do 6!. However, we then have to eliminate the repetitions. We have two sets of 3 items that are the same. So we can eliminate repetition by dividing by (3! × 3!). This accounts for the fact that 3 shirts are an identical red and 3 shirts are an identical green. So we have:

# of outcomes = ` = 20. `

The MISSISSIPPI rule is explained in the Counting with Identical Items lesson linked in the "Related Lessons" section below :)

**FAQ: Why isn't the answer 6! = 6x5x4x3x2x1 =720?**

**A:**6! would be 100% correct if there were

*six unique*different shirts that we were distributing among the six children.

For example, suppose we had six completely different shirts, {A, B, C, D, E, F} -- suppose the kids are #1-6, then

1 = A, 2 = B, 3 = C, 4 = D, 5 = E, 6 = F

would be a different outcome from, say,

1 = B, 2 = A, 3 = C, 4 = D, 5 = E, 6 = F

In the first scenario, Child 1 has shirt A and Child 2 has shirt B. In the second scenario, Child 1 has shirt B and Child 2 has shirt A. If each shirt is unique, then these are two *different* arrangements.

But, in fact, according to the problem, we have "3 identical green shirts and 3 identical red shirts"--- suppose A & B & C are the green shirts and D & E & F are the red shirts --- then the two configurations above would *no longer be different* but rather the* same identical outcome.* In the first scenario both Child 1 and 2 would receive a green shirt, and in the second scenario both children would again receive a green shirt.

Counting as if all six shirts were different leads to major overlap and redundancies, which is why the answer, 720, is A LOT bigger than the answer, 20.

The factorial rule for permutations, n!, only works if all n items are unique and different from one another. If some of the items are the same --- which they are in this case --- we need to use the MISSISSIPPI rule (explained in the Counting with Identical Items lesson video) to remove the redundancies.

**FAQ: Why don't we multiply 6C3 by 2, to get an answer of 40? Shouldn't I do 6C3 for both colors of shirt?**

**A: **Let's start by thinking about a different problem. Change the problem to:

*In how many different ways can 3 identical green shirts be given to 6 children (one child receives one shirt)?*

The answer is 6C3 because we are choosing 3 out of 6 children. This

includes every possible combination of 3 children we can give the 3

green shirts to.

But wait...that's effectively the same as our original question!

Because every combination of 3 children that we DO choose leaves a combination of 3 children that we DIDN'T choose. So we could say this is the same as the above problem, except now the children we didn't choose get the red shirts (instead of nothing).

In other words, every combination of 3 children getting green shirts

includes a combination of the 3 other children getting 3 red

shirts... the 3 leftover children. And since 6C3 covers every possible

combination of children getting the green shirts, we've already

accounted every possible combination. For example:

ABC (green) DEF (red) is included in 6C3

DEF (green) ABC (red) is included in 6C3 also! So we don't want to say

ABC (red) DEF (green) is another combination...because that's the same as DEF (green) ABC (red).

## Related Lessons

Watch the lessons below for more detailed explanations of the concepts tested in this question. And don't worry, you'll be able to return to this answer from the lesson page.