M26-22

Question

In how many ways can we select 8 shirts from a closet containing 7 distinct red shirts and 5 distinct blue shirts, such that at least one shirt of each color remains in the closet, if the order of selection does not matter?

A. 460
B. 490
C. 493
D. 455
E. 445

Show Answer

Official Answer

D

A. 460
B. 490
C. 493
D. 455
E. 445

Bunuel · Accepted Answer

Official Solution:

In how many ways can we select 8 shirts from a closet containing 7 distinct red shirts and 5 distinct blue shirts, such that at least one shirt of each color remains in the closet, if the order of selection does not matter?

A. 460
B. 490
C. 493
D. 455
E. 445

Show Answer

Official Answer

D

A. 460
B. 490
C. 493
D. 455
E. 445

The total number of ways to select 8 shirts out of 12  distinct  shirts (7 red and 5 blue) is given by the combination formula:  C^8_{12} .
 
The number of ways to select 8 shirts such that no red shirts are left in the closet is given by the product of the combinations of choosing all 7 red shirts and 1 blue shirts:  C^7_7 * C^1_5 .
 
The number of ways to select 8 shirts such that no blue shirts are left in the closet is given by the product of the combinations of choosing all 5 blue shirts and 3 red shirts:  C^5_5 * C^3_7 .
 
Therefore, the number of ways to select 8 shirts so that at least one red shirt and at least one blue shirt remain in the closet can be found by subtracting the sum of the two previous cases from the total number of ways to select 8 shirts:  C^8_{12} - (C^7_7 * C^1_5 + C^5_5 * C^3_7) = 495 - (5 + 35) = 455 .

Answer: D

Bunuel · Answer

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

Aishyk97 · Answer

Hi  Bunuel ,
Can you help me identify the gap in my reasoning.

If I were to leave 1 blue shirt and 1 red shirt in the closet.

I will have to choose the 8 shirts from the remaining shirts. That is from 6R and 4B.

But it is not 10C8 or  \frac{10*9*8*7*6*5*4*3}{6!*4!}

I seem to be missing something. Any help is appreciated.

After some thought, I feel I cannot club the red and blue as I may not always choose all 6 reds so I cant calculate it directly.
Is this where I am going wrong?

Bunuel · Answer

Yes.

Apart from that, note that if we want at least one shirt of each color to remain in the closet, selecting 6R and 4B is not the only case. What about 5R + 3B, or 4R + 4B? That's why considering 1- (the probability of the opposite event) is easier here.

Hope it helps.

AnumFatima · Answer

Hi  Bunuel ,

I used the following approach, is it correct? I do realize that it is a longer way.

Total shirts left behind = 12-8 =4

4 left behind can have 1 red and 3 blue
Selected shirts = 7C6 × 5C2 =7 × 10 = 70

4 left behind can have 2 red and 2 blue
Selected shirts = 7C5 × 5C3 = 21 × 10 = 210

4 left behind can have 3 red and 1 blue
Selected shirts = 7C4 × 5C4 = 35 x 5 = 175

Total combinations = 70 + 210 + 175 = 455

Bhavita. · Answer

Hi  Bunuel ,

Since we need to ensure at least one shirt of each color remains in the closet
7C1 * 5C1 among 7 distinct and 5 distinct diff color shirts 
and then for the rest 10 we have 10C8 ways to choose

the answer comes out to be 7*5*45

Could you please explain me where exactly I am going wrong

Bunuel · Answer

This method results in duplicate sets of 8 shirts.

For example, one set might include Red 1 and Blue 1, selected using 7C1 * 5C1, along with the remaining 6 shirts determined by 10C6, such as Red 2, 3, 4, 5, 6, 7.

However, another possible set could include Red 2 and Blue 1, also selected using 7C1 * 5C1, alongside Red 1, 3, 4, 5, 6, 7.

These two sets are identical: {Red 1, 2, 3, 4, 5, 6, 7, Blue 1}.

sb995 · Answer

I did not quite understand the solution. Hi, could you please help explain the gap in my reasoning below:

Selecting 6 red shirts and two blue shirts: (7C6)*(5C2) +

Selecting 4 red shirts and 4 blue shirts: (5C4)*(7C4)

Total combinations: (7C6)*(5C2) + (5C4)*(7C4).

But this is less than the official answer

Bunuel · Answer

The gap in the reasoning is that you only listed two distributions (6 red + 2 blue and 4 red + 4 blue). But the problem requires at least one shirt of each color left in the closet, so multiple other valid splits exist, such as 5 red + 3 blue or 3 red + 5 blue.

That’s why the official solution works: it counts all 8-shirt selections first and then subtracts the cases where all reds are taken or all blues are taken. This way, every valid distribution that leaves at least one of each color is included, not just the two you considered.

You will always miss the question and fail to understand the solutions if you do not read the question correctly. If you do not read it correctly, then by definition you are solving the wrong question.

chrwens309 · Answer

I like the solution - it’s helpful.

PriChats · Answer

To simply visualize this problem, think of the   = FAVORABLE_CASES

Total cases = 12C8 to pick 8 shirts out of 12 shirts "

There are 2 things that could have happened when the question says that "at least one shirt of each color remains in the closet":
1. 0 shirt remains
2. Only 1 shirt remains

Because if we leave 1 shirt per color, there are in total 2 shirts remaining in the closet. Therefore only 2 negatives cases exists.

Now we remove the unfavorable cases from the total. 
1st unfavorable case - 0 shirts remaining : we have 0 cases as what we are taking is always less than the total. we are picking 8 out of 12 total. Hence, there would never be a case where 0 shirt remains.
2nd unfavorable case - 1 shirt remaining : Here things get interesting, We can pick which colored shirt can remain - BLUE or RED (Mind it all the shirts are DISTINCT!!)

FAVORABLE_CASES = TOTAL - UNFAVORABLE_CASES => 12C8 - 0 - 5C5.7C3 (Leaving Blue) - 5C1.7C7 (Leaving Red) = 495 - 0 - 35 - 5 = 455

Hence, the solution is 455

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