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RivalryofTroll

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In how many ways can 5 men and 5 women be arranged in a circle so that the men are separated?

I got 2880 so 5! x 4!

Continuation: In how many ways can this be done if two particular women must NOT be next to a particular man?

I thought it was something like: 2880 - (the ways where the two particular women must be next to a particular man) but yeah, not sure how to do this.
 

Carrotsticks

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A little ambiguous.

Let the women be A and B, and let the man be M.

So obviously AMB is not allowed, but how about ABM or MAB etc?
 

RivalryofTroll

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A little ambiguous.

Let the women be A and B, and let the man be M.

So obviously AMB is not allowed, but how about ABM or MAB etc?
yeah that's allowed.

The answer is 864 if you know how to get that...
 

RealiseNothing

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Place the particular man, then arrange the other 4 men, which can be done in 4! ways.

Now there are 5 spots left for the women, 2 of which are next to the particular man. Place the 2 women who can't sit next to the man in two of the three "other" seats, to get 3x2= 6.

Now place the remaining 3 women in the last 3 seats = 3! = 6.

Therefore answer is 4! x 6 x 6 = 36 x 24 = 30^2 - 6^2 = 900 - 36 = 864.
 

RivalryofTroll

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Place the particular man, then arrange the other 4 men, which can be done in 4! ways.

Now there are 5 spots left for the women, 2 of which are next to the particular man. Place the 2 women who can't sit next to the man in two of the three "other" seats, to get 3x2= 6.

Now place the remaining 3 women in the last 3 seats = 3! = 6.

Therefore answer is 4! x 6 x 6 = 36 x 24 = 30^2 - 6^2 = 900 - 36 = 864.
fanks.
 

theind1996

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Place the particular man, then arrange the other 4 men, which can be done in 4! ways.

Now there are 5 spots left for the women, 2 of which are next to the particular man. Place the 2 women who can't sit next to the man in two of the three "other" seats, to get 3x2= 6.

Now place the remaining 3 women in the last 3 seats = 3! = 6.

Therefore answer is 4! x 6 x 6 = 36 x 24 = 30^2 - 6^2 = 900 - 36 = 864.
What if there is a male sitting next to "the particular man"?
 

RivalryofTroll

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A committee of 6 is to be selected from 10 people of whom A and B are two. How many committees can be formed excluding A if B is included?
 

Carrotsticks

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A committee of 6 is to be selected from 10 people of whom A and B are two. How many committees can be formed excluding A if B is included?
If they are to be included, then we only need to pick 4 people now, instead of 6.

But since A and B are already in it, we only have 8 to choose from, instead of 10.

So 8C4.
 

RivalryofTroll

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If they are to be included, then we only need to pick 4 people now, instead of 6.

But since A and B are already in it, we only have 8 to choose from, instead of 10.

So 8C4.
Ummm, éxcluding A if B is included?
(read question again)
 

RivalryofTroll

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If B is included, then there are 5 others to be selected from 8 possible ones (since A can't be there), so 8C5?
I did 8C5 = 56 but answers say 140.

Need to confirm if I'm wrong or answers are wrong.
 

Carrotsticks

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Ummm, éxcluding A if B is included?
(read question again)
Oh misread question.

In that case, we have 8 people to choose from now since B is already in there and A is NOT allowed to be chosen.

And 5 spots, since B is already in there.

So 8C5.
 

theind1996

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For circular ones, what do we do when one person won't sit next to another? (let's say out of 10 people)
 

Carrotsticks

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For circular ones, what do we do when one person won't sit next to another? (let's say out of 10 people)
Suppose A and B are in a circle, and DON'T want to sit next to each other.

Then we do Total - together = 9! - 8!2!
 

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