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 January 24th, 2017, 07:33 PM #1 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 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.
 January 24th, 2017, 07:58 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,571 Thanks: 1415 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.
 January 24th, 2017, 11:29 PM #3 Senior Member   Joined: Jul 2011 Posts: 405 Thanks: 16 Thanks romsek answer given as $960$

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