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
PS_Counting
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- vittovangind
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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?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
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
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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.
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.
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.
Let GGG = the 3 identical green shirts and RRR = the 3 identical red shirts.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
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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- vittovangind
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Brent,Brent@GMATPrepNow wrote: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?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
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
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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Fair enough.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
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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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.
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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The key word here is "identical"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.
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.
Cheers,
Brent
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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
GMAT assassins aren't born, they're made,
Rich
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
GMAT assassins aren't born, they're made,
Rich
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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
is 6! / (3! * 3!) = 6 * 5 * 4 * 3! / 6 * 3! = 5*4 = 20