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In how many different ways can 3 identical green shirts and
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Updated on: 24 Feb 2012, 22:58

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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?

Re: In how many different ways can 3 identical green shirts and
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24 Feb 2012, 22:59

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shrive555 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?

So, basically # of assignments of 6 shirts to 6 children (such that each child receives a shirt) equals to # of permutations of 6 letters BBBGGG, which is \(\frac{6!}{3!3!}=20\) (we divide by 3!*3!, since there are 3 identical B's and 3 identical G's).

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?

20 40 216 720 729

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.
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Re: In how many different ways can 3 identical green shirts and
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30 Mar 2015, 01:55

Temurkhon wrote:

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?

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
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Re: In how many different ways can 3 identical green shirts and
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29 Aug 2016, 14:14

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shrive555 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 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)

Re: In how many different ways can 3 identical green shirts and
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29 Aug 2016, 19:15

shrive555 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

6! / 3!x3! = 120/6= 20 ways
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Re: In how many different ways can 3 identical green shirts and
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29 Nov 2016, 16:28

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shrive555 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

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:

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:

Re: In how many different ways can 3 identical green shirts and
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13 Feb 2018, 22:58

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.

Re: In how many different ways can 3 identical green shirts and
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04 Nov 2019, 05:40

shrive555 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

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

In how many different ways can 3 identical green shirts and
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04 Nov 2019, 05:49

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?

Re: In how many different ways can 3 identical green shirts and
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04 Nov 2019, 05:58

tinytiger wrote:

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

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