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In how many different ways can 3 identical green shirts and

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In how many different ways can 3 identical green shirts and  [#permalink]

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Updated on: 24 Feb 2012, 22:58
3
12
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Difficulty:

35% (medium)

Question Stats:

66% (01:16) correct 34% (01:36) wrong based on 288 sessions

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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

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Originally posted by shrive555 on 25 Nov 2010, 18:54.
Last edited by Bunuel on 24 Feb 2012, 22:58, edited 1 time in total.
Edited the question
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25 Nov 2010, 19:51
2
2
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
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25 Nov 2010, 20:01
No of ways 6 shirts can be distributed among 6 people 6!
Since 3 red are identical and 3 green are identical = (6!)/3!*3!=20
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25 Nov 2010, 21:43
2
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?

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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24 Feb 2012, 17:19
Approach 1:
1st Child: 6 has options
2nd Child: 5 has options…
Therefore, for all kids: 6 x 5 x 4 x 3 x 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, 720/[ 3! X 3! ] = 20 ways.

Approach 2 / MGMAT technique:
This is like anagramming RRRGGG. No of arrangements = 6! / 3! x 3! ways ==> 20.
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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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24 Feb 2012, 22:59
3
1
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

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 # 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).

Hope it helps.
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In how many different ways can 3 identical green shirts and  [#permalink]

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30 Mar 2015, 01:18
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?
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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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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  [#permalink]

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30 Mar 2015, 02:07
Karishma,

thanks for immediate response. I'm confined by one concept
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In how many different ways can 3 identical green shirts and  [#permalink]

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30 Mar 2015, 03:36
What about 6 shirts and 5 children?

My view is that we have two options to be distributed:

3 Red shirts and 2 Green shirts

OR

2 Red shirts and 3 Green shirts

5!/3!*2!=10*2=20

What do you think, Karishma. I'm sorry to be annoying
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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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29 Aug 2016, 14:14
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Top Contributor
1
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)

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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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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  [#permalink]

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17 Nov 2016, 19:39
6!/3!3! = 20 --> Combination that takes into account two items (in this case, shirts), that are identical.

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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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29 Nov 2016, 16:28
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:

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

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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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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.

720/6 = 120
120/6 = 20

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Re: In how many different ways can 3 identical green shirts and  [#permalink]

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Re: In how many different ways can 3 identical green shirts and   [#permalink] 10 Mar 2019, 05:38
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