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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?

Randhawa
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  • 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. – Abr001am Dec 03 '17 at 21:27
  • I got this idea by watching other question on stack but still not able to find the answer. – Randhawa Dec 03 '17 at 21:28
  • can you try at least ? – Abr001am Dec 03 '17 at 21:29
  • Yes i m trying from previous two hours :( – Randhawa Dec 03 '17 at 21:30
  • Are these separate questions? If so, you should read the answers to this question, which will help you with the first question. – N. F. Taussig Dec 03 '17 at 22:04
  • This is one statement and then there are two questions arises from that statement. I can solve easily if there are two groups( men ,women) or (children, adults) but there are three ( couples,adults, children )so i am not able to find any solution. – Randhawa Dec 03 '17 at 22:10

1 Answers1

2

a) $9! \cdot 10C3 \cdot 3!$

b) $9! \cdot 2! \cdot 2! \cdot 2!$

amWhy
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Randhawa
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  • Your answers are correct. – N. F. Taussig Dec 04 '17 at 00:06
  • I know that , I confirmed it from professor. But I am still confused how i got answer of b part – Randhawa Dec 04 '17 at 02:21
  • 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. – N. F. Taussig Dec 04 '17 at 03:16