In how many different ways can 3 identical green shirts and

Question

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

Bunuel · Accepted Answer

1-2-3-4-5-6  (children)
B-B-B-G-G-G (shirts)
G-B-B-G-G-B
G-G-B-G-B-B
....

So, basically, the number of ways to assign 6 shirts to 6 children (such that each child receives a shirt) equals the number of permutations of the 6 letters BBBGGG, which is 6!/(3! * 3!) = 20 (we divide by 3! * 3! because there are 3 identical B's and 3 identical G's).

Answer: A.

Hope it helps.

LM · Answer

GGG RRR

Therefore total number of ways is

6! but there are two groups of 3 identical things.

Therefore total number of "different" ways is

6!/ (3!) (3!) = 20

KarishmaB · Answer

Or out of 6 children, choose 3 in 6C3 ways = 20 ways.

Note: When you choose 3 children say, A, B and C are give them a red shirt, D, E and F get a green shirt. When you choose D, E and F and give them a red shirt, A, B and C automatically get the green shirts. So you do not need to multiply by 2! above.

Temurkhon · Answer

What if we distribute 6 shirts among 4 children?

R-R-R-G-G-G (shirts)
1-2-3-4 (children)

RRRG
RRGG
RGGG
GGGR
GGRR
GRRR
RGRG
GRGR
RGGR
GRRG

Looks like 10

algebraically 6*5*4*3/3! *3!=10

Is that right?

KarishmaB · Answer

You mean each child gets exactly one shirt?

You have missed 4 cases: GRGG, GGRG, RGRR, RRGR

There will be total 14 cases. 
Say, had there been 4 of each type of shirt, each child could have got a shirt in two ways: Red or Green. This would give us 2*2*2*2 = 16 ways.
But "All 4 children get Red" and "All 4 children get Green" are two cases which are not possible. So total cases are 16 - 2 = 14

BrentGMATPrepNow · Answer

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

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ScottTargetTestPrep · Answer

If we let G denote a green shirt and R denote a red shirt, the problem becomes how to arrange 3 Gs and 3 Rs in the string of GGGRRR. The answer can be found using the concept of permutations with repetition of indistinguishable objects, using the following formula:

https://i.imgur.com/ll2sz2e.png

In this formula, N represents the total number of objects to be arranged. Each ri (i = 1, 2, 3, …, n) represents the frequency of each of the i indistinguishable objects.

The frequency simply means the number of times that the indistinguishable item occurs in the set. Note that there are 3 green shirts that are identical (indistinguishable), and there are 3 red shirts that are identical (indistinguishable).
 
Therefore, the number of ways we can arrange 3 Gs and 3 Rs in the string of GGGRRR is:

6!/(3! × 3!) = 720/(6 × 6) = 720/36 = 20

Answer: A

EMPOWERgmatRichC · Answer

Hi All,

If we had 6 different shirts, then there would be 6! = 720 options. However, since there are 3 IDENTICAL red shirts and 3 IDENTICAL green shirts, we have to do some extra work on our calculations.

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

GMATGuruNY · Answer

From 6 children, the number of ways to choose 3 to receive green shirts = 6C3 = (6*5*4)/(3*2*1) = 20
From the 3 remaining children, the number of ways to choose 3 to receive red shirts = 3C3 = (3*2*1)/(3*2*1) = 1
To combine these options, we multiply:
20*1 = 20

A .

tinytiger · Answer

Hi experts, we’re assuming that the 6 children are not distinct here. In the case they are distinct, may I ask how to approach the question in this case?

Do we have to multiply the answer of 20 by the number of distinct arrangements from the distinct kids?

Posted from my mobile device

GMATGuruNY · Answer

The children are distinct.
If the 6 children are Adam, Bobby, Cindy, David, Ellen, and Frank, clearly no two children are the same.

GMATmona07 · Answer

Hi  karishma   Bunuel ,
I've a doubt. Why shouldn't we consider 2 cases(one green or one red) for each child, giving a total 2^6 no. of combinations?

EMPOWERgmatRichC · Answer

Hi GMATmona07, In the situation that you are calculating, you could potentially have all 6 children with the same color shirt (either all reds or all greens), but that is NOT what the prompt describes to us. There are only 3 of each color (meaning that 3 children will end up with a red shirt and 3 will end up with a green shirt) AND each set of three shirts is identical - so we have to account for those details in our calculation. GMAT assassins aren't born, they're made, Rich Contact Rich at: Rich.C@empowergmat.com

kittle · Answer

Hey KarishmaB - why do we not divide the below by 2? The way we do in the question on this link: https://gmatclub.com/forum/how-many-way ... 05381.html In both the questions, we are trying to group (find different ways) I am unable to wrap my head around this. Please help me KarishmaB Your explanations will really help me improve upon this topic.

KarishmaB · Answer

Here, in this question, we are splitting 6 people into two distinct groups - red shirt and green shirt groups.

Whereas, we divide by 2! in that question because the teams are not different. We are just putting them into two groups. There is no team 1 and team 2 in that question.

djangobackend · Answer

will we not arrange them? isn't this just selection?

Bunuel · Answer

The arrangement isn't necessary here because the shirts are identical within each color. We're simply choosing which 3 out of the 6 children get green shirts. The remaining 3 children will automatically receive red shirts. Check another solution here: https://gmatclub.com/forum/in-how-many- ... l#p1049714 Hope it helps.

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In how many different ways can 3 identical green shirts and

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