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# Combinations with duplicates! watch

1. Sean and Hana are going to bring 8 books with them. Sean has 16 books to choose from, and Hana has 24 books to choose from.

If Hana and Sean EACH pick 4 out of 40 books, and some of them were duplicates, how would you calculate the probability of this? (By duplicate, Sean and Hana both have a copy of the same book).

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My friend is saying that this can be done by

40!/8!*32!*2!, where 2! will eliminate the duplicate. But I don't think this is true???
2. (Original post by FuryBall)
Sean and Hana are going to bring 8 books with them. Sean has 16 books to choose from, and Hana has 24 books to choose from.

If Hana and Sean each pick 4 out of 40 books, and some of them were duplicates, how would you calculate the probability of this? (By duplicate, Sean and Hana both have a copy of the same book).

----------------------

My friend is saying that this can be done by

40!/8!*32!*2!, where 2! will eliminate the duplicate. But I don't think this is true???
There seems to be some confusion, even before you get to the duplicates.

From the first lines I get the impression Sean chooses 4 from his 16, and Hana chooses 4 from her 24. BUT then you say they choose 4 each from 40, which is a different situation.

3. (Original post by ghostwalker)
There seems to be some confusion, even before you get to the duplicates.

From the first lines I get the impression Sean chooses 4 from his 16, and Hana chooses 4 from her 24. BUT then you say they choose 4 each from 40, which is a different situation.

I mean that they EACH choose 4 from 40, so together it is still 8 books picked from 40.
4. (Original post by FuryBall)
I mean that they EACH choose 4 from 40, so together it is still 8 books picked from 40.
And what are you trying to calculate? The probability that they choose a duplicate book? Or is it the number of choices they can make in the 8, which seems to be what your friend is working on? The value is going to depend on the number of books that are duplicated as well - is it just one, or more than one?
5. (Original post by ghostwalker)
And what are you trying to calculate? The probability that they choose a duplicate book? Or is it the number of choices they can make in the 8, which seems to be what your friend is working on? The value is going to depend on the number of books that are duplicated as well - is it just one, or more than one?
The question from my project doesn't specify the number of duplicates as it just says "some duplicates". But I could specify it as 1 duplicate and then find a way of applying my method to x number of duplicates.

I think I have to find the way they can choose the books while eliminating/ignoring the probability of a duplicate.
I don't understand what my friend's calculation is exactly doing.
6. (Original post by FuryBall)
The question from my project doesn't specify the number of duplicates as it just says "some duplicates". But I could specify it as 1 duplicate and then find a way of applying my method to x number of duplicates.

I think I have to find the way they can choose the books while eliminating/ignoring the probability of a duplicate.
I don't understand what my friend's calculation is exactly doing.

Well your friends calculation is incorrect. You can safely divide by 2!, if each choice contained the two duplicates, but it doesn't; some choices will have one of the duplicate pair and some will have none.

This gives us the clue on how to break this down.

Either both duplicates are chosen, so we're left with how many ways of choosing the remaining 6 from the remaining 38; which I presume you're fine with.

Or one of the duplicates is chosen, so we then have to choose 7 from 38 (since we can't choose the 2nd duplicate).

Or neither duplicate is chosen, and we just choose 8 from 38.
7. (Original post by ghostwalker)
Well your friends calculation is incorrect. You can safely divide by 2!, if each choice contained the two duplicates, but it doesn't; some choices will have one of the duplicate pair and some will have none.

This gives us the clue on how to break this down.

Either both duplicates are chosen, so we're left with how many ways of choosing the remaining 6 from the remaining 38; which I presume you're fine with.

Or one of the duplicates is chosen, so we then have to choose 7 from 38 (since we can't choose the 2nd duplicate).

Or neither duplicate is chosen, and we just choose 8 from 38.
Interesting...

But if we have A A B C, the internal ordering of the duplicate A is 2!, and therefore 4!/2!*2!*2! should be correct? (2! in the case for each duplicate set. Divide by 2! * 2! for 2 titles of duplicates (I imagine something like A A B B C), and so on.

So could we say

40!(8!)(32!)(2!)^d with d being the set of duplicates.

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