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Three single men, two single women and two families take their places at a round table. Each of the two families consists of two parent and one child. Find the number of possible seating arrangements.

  1. members of the same family are seated together and the two single women are separated
  2. the seats are numbered and each child sits between their parents

My attempts,

  1. $\frac{6!}{6} + 3!\cdot2\cdot6!=87604$
    Correct Answer:$17280$
  2. $\frac{8!}{8}\cdot\ _3C_2\cdot2!=30240$
    Correct Answer:$31680$
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For $1$, each family forms a block. There are $7!/7$ inequivalent ways to arrange the $7$ men, women and blocks. With the single women also forming a block, there are $6!/6$ inequivalent ways. In the first case, there are $3!^2$ internal permutations for the families, and in the second case there are $3!^2$ internal permutations for the families and $2$ for the single women. Thus the total number of inequivalent admissible arrangements is

$$ (3!^2)7!/7-2(3!^2)6!/6=17280\;. $$

If the seats are numbered and each child sits between their parents, there are $11\cdot7!/7$ arrangements of the men, women and family blocks, and $2^2$ internal arrangements for the two families, yielding

$$ 2^2\cdot11\cdot7!/7=31680 $$

different arrangements.

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  • $\begingroup$ Thanks for answering my question! help I haven't learnt about internal permutations yet, could you please help me break it down , what is internal permutations? $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 18:27
  • $\begingroup$ hmm i understand now, thanks XD the 3!^2=3 *2*1 * 3*2*1 is because there's 2 families so need to count twice, ok ! i got it! thanks! $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 18:30
  • $\begingroup$ But for qns 2 why need to multiply by 11 >< i don't understand, help me!! $\endgroup$ – DreadfulWithMaths Feb 4 '16 at 18:35
  • $\begingroup$ @DreadfulWithMaths: There are $7!/7$ cyclically inequivalent arrangements of the $7$ men, women and family blocks. There are $11$ seats, so each of these cyclically inequivalent arrangements can be placed on the numbered seats in $11$ different ways. $\endgroup$ – joriki Feb 4 '16 at 18:53
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Find the number of seating arrangements if members of the same family are seated together and the two single women are separated.

Seat one of the women. We will position everybody else relative to her.

Once that woman has been seated, we are left with six objects (the three single men, two families, and the other single woman) to arrange as we move clockwise around the circle. We can arrange the objects in $6!$ ways as we move around the circle (place an object to the first woman's immediate left, the next object to the immediate left of that object, and so forth). We can arrange the three members of each family internally in $3!$ ways. Hence, if there were no restrictions on the placement of the two single women, there would be $6! \cdot 3!^2$ arrangements of the three single men, two single women, and two families around the table.

From these arrangements, we must exclude those in which the two single women sit in adjacent seats. Sit one of the women. If the other women sits next to her, there are two ways to seat her, to the immediate left or the immediate right of the first women. Once the women have been seated, there are five objects (the three single men and two families) left to arrange. These objects can be arranged in $5!$ ways as we move clockwise around the circle from the block of two single women. The two families can each be arranged internally in $3!$ ways. Hence, the number of arrangements in which the two single women sit in adjacent seats is $2 \cdot 5! \cdot 3!^2$.

Therefore, the number of seating arrangements in which members of the same family are seated together and the two single women are separated is $$6! \cdot 3!^2 - 2 \cdot 5! \cdot 3!^2$$

Find the number of seating arrangements if the seats are numbered and each child sits between his or her parents.

Again, we seat one of the women, then position everyone else relative to her.

Since there are eleven people, there are $11$ ways to seat the aforementioned woman in one of the seats. This leaves us with six objects to arrange (the three single men, the two families, and the other single woman). They can be arranged in $6!$ ways as we move clockwise around the circle. Since each child must be seated between his or her parents, each family can be arranged internally in $2!$ ways (only the parents can switch places). Hence, the number of seating arrangements if the seats are numbered and each child sits between his or her parents is $$11 \cdot 6! \cdot 2!^2$$

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