In how many different ways can 3 identical green shirts and

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

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.

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.

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).

Answer: A.

Hope it helps.

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

Karishma,

thanks for immediate response. I'm confined by one concept

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

6! / 3!x3! = 120/6= 20 ways

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

A.

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

Answer: A

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:

GMAT assassins aren't born, they're made,
Rich

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

.

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.

Yes you’re right - I missed that out. Thanks again for clarifying in such short notice!

Posted from my mobile device

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?