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by vittovangind » Sat Jan 25, 2014 9:34 am
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
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by Brent@GMATPrepNow » Sat Jan 25, 2014 9:42 am
vittovangind 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

By the way, if anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

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by GMATGuruNY » Sat Jan 25, 2014 1:08 pm
Alternate approach:

Number of ways to arrange n distinct elements = n!.
When an arrangement includes IDENTICAL elements, we must DIVIDE by the number of ways each set of identical elements can be arranged.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Thus:
Number of ways to arrange AAABB = 5!/(3!2!).
We divide by 3! to account of the 3 identical A's and by 2! to account for the 2 identical B's.
vittovangind 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
Let GGG = the 3 identical green shirts and RRR = the 3 identical red shirts.
Every unique arrangement of the 6 letters GGGRRR represents one way to give each child a shirt.
Number of ways to arrange GGGRRR = 6!/(3!3!) = 20.

The correct answer is A.
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by vittovangind » Sun Jan 26, 2014 1:21 pm
Brent@GMATPrepNow wrote:
vittovangind 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

By the way, if anyone is interested, we have a free video on calculating combinations (like 6C3) in your head: https://www.gmatprepnow.com/module/gmat-counting?id=789

Cheers,
Brent
Brent,

Many thanks for your feedback. My issue still is that I'm having a hard time reformulating the question as you did.

Cheers,

Vitto
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by Brent@GMATPrepNow » Sun Jan 26, 2014 1:31 pm
vittovangind wrote: Brent,

Many thanks for your feedback. My issue still is that I'm having a hard time reformulating the question as you did.

Cheers,

Vitto
Fair enough.
Here's a similar solution that utilizes the Fundamental Counting Principle (perhaps a more familiar approach).

Take the task of seating handing out shirts and break it into stages.

Aside: When breaking a task into stages, you should ask yourself "How would I accomplish this task?" Here's one such way:

Stage 1: hand out the 3 green shirts
So, we must 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)

Stage 2: hand out the 3 red shirts
Now that we have given green shirts to 3 of the 6 children, the remaining 3 children MUST receive the red shirts.
So, there's only 1 way to hand out the red shirts.

By the Fundamental Counting Principle (FCP), we can complete the two stages (and thus hand out the 6 shirts) in (20)(1) ways ([spoiler]= 20 ways = A[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, watch our free video: https://www.gmatprepnow.com/module/gmat-counting?id=775
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by j_shreyans » Fri May 15, 2015 8:01 am
Hi Brent ,

According to the question among 6 children such that each child receives the shirt.

SO can't we do like below.

1st can receive the shirt in 6 ways

2nd can receive the shirt in 5 ways

3rd can receive the shirt in 4 ways

4th can receive the shit in 3 ways

5th can receive the shirt in 2 ways

6th can receive the shirt in 1 ways

so 6X54X3X2X1=720.

Please correct me if i am wrong.
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by Brent@GMATPrepNow » Fri May 15, 2015 8:07 am
j_shreyans wrote:Hi Brent ,

According to the question among 6 children such that each child receives the shirt.

SO can't we do like below.

1st can receive the shirt in 6 ways

2nd can receive the shirt in 5 ways

3rd can receive the shirt in 4 ways

4th can receive the shit in 3 ways

5th can receive the shirt in 2 ways

6th can receive the shirt in 1 ways

so 6X54X3X2X1=720.

Please correct me if i am wrong.
The key word here is "identical"
We have 3 identical green shirts and 3 identical red shirts.
So, the 1st child can receive EITHER a green shirt or a red shirt (2 options)
And so on.

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by [email protected] » Fri May 15, 2015 9:02 am
Hi j_shreyans,

Since there are 3 IDENTICAL red shirts and 3 IDENTICAL green shirts, you would have to do one more step in your calculation.

If we call 3 of the children A, B and C and each of them gets an identical red shirt, then there are technically 6 different ways for those 3 shirts to be given to those 3 children. Since the shirts are identical though, we are NOT supposed to count this as 6 different options....it should only be counted as 1 option.

Thus, we would have to divide 720 by 6....

We would have to do the same thing with the identical green shirts, which means we'd have to divide by 6 AGAIN.

720/6 = 120
120/6 = 20

Final Answer: A

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by nikhilgmat31 » Wed Oct 07, 2015 2:31 am
Simple way to remember such question is number of way to arrange GGGRRR

is 6! / (3! * 3!) = 6 * 5 * 4 * 3! / 6 * 3! = 5*4 = 20
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