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January 24th, 2017, 07:33 PM   #1
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circular arrangement

The number of ways in which $4$ couples can be seated around a round table

if one particular couple always sit together but one particular couple never together.

**Attempt** first we will select one particular couple is $\displaystyle \binom{4}{1}$

now select other particular couple couple is $\displaystyle \binom{3}{1}$

Now how can I solve it? Help me, Thanks.

Last edited by skipjack; January 25th, 2017 at 01:15 AM.
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January 24th, 2017, 07:58 PM   #2
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Let's assume that rotations of the entire arrangement are indistinguishable.

Place your initial couple. There are 2 ways of seating them together.

This leaves 6 seats.

Take one of the people from the couple that won't sit together,

If they sit in one of the seats adjacent the first couple, then there are 4 ways to sit the other member of that couple. There are 2 of these seats.

If they sit in one of the other 4 seats, there are 3 ways to sit the other member of that couple.

This gets you a total of $2(2\cdot 4 + 4 \cdot 3) = 40$ ways to seat these two couples.

Then there are 4 seats left and $4!=24$ ways to seat the rest.

Thus there are a total of $40\cdot 24 = 960$ ways to seat the 4 couples.

If seats are labeled and rotations of the entire arrangement are distinguishable, this adds another factor of 8 and thus there become 7680 arrangements.
Thanks from panky

Last edited by skipjack; January 25th, 2017 at 01:16 AM.
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January 24th, 2017, 11:29 PM   #3
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Thanks romsek answer given as $960$
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