There are twelve people which includes 3 couples, 3 single adults and 3 children. In how many ways they can be arranged :-
a) if no two children can sit in adjacent seats?
b) if each couple must sit in adjacent seats?
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$\begingroup$ for a) try to find arrangements of 9 elements instead of 12, because 3 of them are acually couples of a child and a random person from the rest, b) same thing just the couples are not random. $\endgroup$– Abr001amCommented Dec 3, 2017 at 21:27
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$\begingroup$ I got this idea by watching other question on stack but still not able to find the answer. $\endgroup$– RandhawaCommented Dec 3, 2017 at 21:28
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$\begingroup$ can you try at least ? $\endgroup$– Abr001amCommented Dec 3, 2017 at 21:29
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$\begingroup$ Yes i m trying from previous two hours :( $\endgroup$– RandhawaCommented Dec 3, 2017 at 21:30
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$\begingroup$ Are these separate questions? If so, you should read the answers to this question, which will help you with the first question. $\endgroup$– N. F. TaussigCommented Dec 3, 2017 at 22:04
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1 Answer
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a) $9! \cdot 10C3 \cdot 3!$
b) $9! \cdot 2! \cdot 2! \cdot 2!$
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$\begingroup$ I know that , I confirmed it from professor. But I am still confused how i got answer of b part $\endgroup$– RandhawaCommented Dec 4, 2017 at 2:21
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$\begingroup$ Treat each couple as a single object. Then you have nine objects to arrange, the three couples, the three single adults, and the three children. You can arrange the objects in $9!$ ways. Each of the three couples can be arranged internally in $2!$ ways. Hence, there are $9! \cdot 2!^3$ arrangements in which the couples sit in adjacent seats. $\endgroup$ Commented Dec 4, 2017 at 3:16