In a building, 6 men and 4 women are living. There are 3 married couples. How many ways who the couples are can be guessed?

## 5 Answers

Bad video. The more straightforward way to look at it is

Pick 3 men from 6

Pick 3 women from 4

How many ways to pair these 3 men and 3 women

Multiply it all together to get the answer
I'm sorry, but that video is completely unintelligible; the quality is horrible and I have no idea what the lecturer is saying (I say as someone who's Indian). I couldn't really follow along with his proof, either.

Lets assume that these are all straight marriages. That means that 3 of the 4 women are married to 3 of the 6 men.

Choose one woman out of 4, and choose one man out of 6. They are now married and accounted for 4x6 = 24 possibilities. Then choose another woman out of the remaining 3 women. She is married to one of the 5 remaining men. There are 3 x 5 = 15 ways to do this, so we now have a total of 15 x 24 = 360 choices. Then choose another woman from from the remaining 2 women, and assign her one of the 4 remaining men. That yields 2 x 4 = 8 possibilities, so there are 8 x 360 = 2880 possible couples. This matches with what you calculated I believe (though Reddit seemed to have formatted what you said incorrectly).

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Edit: In fact, you need to divide by 6 to account for the ordering possibilities, as you could have chosen the 3 couples in different orders, and we need to avoid double counting. There are 6 ways to order 3 things (3 choices of the first thing, 2 choices of the second thing, and 1 choice for the last thing to determine the ordering, so 3x2x1 = 6 choices of ordering). Thus, the correct answer is 2880/6 = 480, which is 6x5x4x4. I still maintain that the proof given in the video was impossible to follow, though it does yield the correct answer.
You don't want to double-count, or actually sextuple count.

Suppose these are male-female couples.

If you said this was 6 x 5 x 4 men and 4 x 3 x 2 women, you could get this scenario

m1w1 m2w2 m3w3

But also

m3w3 m2w2 m1w1

The same pairs chosen in a different order.
First select the woman who is not married. There are 4 choices.
Next, write down the names of the three other women in alphabetical order.
The first woman could be married to one of 6 men. Then the second to one of 5 men, and the third to one of 4.
So there are 4.6.5.4 = 480 possibilities.
When I solved it by pausing the video immediately after reading the question I've arrived at the answer 2880, but watching the video for the answer left me super confused. Thank you for explaining, much appreciated.

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