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In how many different ways can 3 identical green shirts and

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by BTGmoderatorDC » Thu Jul 11, 2019 9:37 pm

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Answers

A

B

C

D

E

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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

OA A

Source: Magoosh
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by Jay@ManhattanReview » Thu Jul 11, 2019 10:00 pm
BTGmoderatorDC 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

OA A

Source: Magoosh
If there are p numbers of identical items, q numbers of identical items, and r numbers of identical items, the total number of way, (p + q + r) items be distributed is given by (p + q + r)! / (p!*q!*r!).

Thus, the number of ways 3 identical green shirts and 3 identical red shirts be distributed among 6 children = 6! / 3!*3! = 20

The correct answer: A

Hope this helps!

-Jay
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by Brent@GMATPrepNow » Fri Jul 12, 2019 5:05 am
BTGmoderatorDC 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
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 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
Brent Hanneson - Creator of GMATPrepNow.com
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by AAPL » Fri Jul 12, 2019 11:04 am
1st Child: 6 has options
2nd Child: 5 has options...
Therefore, for all kids: \(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 = 720\) arrangements.

Since the reds are identical, we divide by \(3!\); Since the greens are identical, we divide by another \(3!\)

So: in all, \(\frac{720}{3! \cdot 3!} = 20\) ways.
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