ktrout2020 wrote: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?

A. 20
B. 40
C. 216
D. 720
E. 729

Source: Magoosh

This is a question on arranging objects of which few of them are indistinguishable/identical objects.

Say there are a total of n objects; out of which there are p identical objects, q identical objects, and r identical objects.

Thus, the no, of ways, n objects can be arranged = n! / (p!*q!*r!)

Let's switch to the question:

There are 6 shirts; thus, n = 6; out of 6 shirts, there are 3 identical green shirts; thus, p = 3. Similarly, there are 3 identical red shirts, q = 3.

Thus, the no. of ways 6 shirts can be distributed = 6! / (3!*3!) = 20.

The correct answer: A

Hope this helps!

-Jay
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ktrout2020 wrote: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?

A. 20
B. 40
C. 216
D. 720
E. 729

Source: Magoosh

We can take this question and ask an easier question: In how many ways can we choose 3 of the 6 children to receive a green shirt?

Notice that, once we have given a green shirt to each of those 3 chosen children, the REMAINING 3 children must get red shirts. In other words, once we have given green shirts to 3 children, the children who get red shirts is locked.

So, in how many ways can we select 3 of the 6 children to receive a green shirt?
Since the order of the selected children does not matter, this is a combination question.
We can choose 3 children from 6 children in 6C3 ways (= 20 ways)

Answer: A

Cheers,
Brent

ktrout2020 wrote: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?

A. 20
B. 40
C. 216
D. 720
E. 729

Source: Magoosh

The number of ways to distribute the shirts to the children is given by the number of arrangements of:

G - G - G - R - R - R

Since we have 6 total letters and 3 repeated G's and 3 repeated R's, we can arrange the letters in the following number of ways, using the indistinguishable permutations formula:

6!/(3! x 3!) = (6 x 5 x 4)/(3 x 2) = 5 x 4 = 20.

Answer: A