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

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.
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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
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Bhavita.
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
­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}.­
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