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In how many ways $4$ men and $3$ women and $5$ children be arranged in a circle if all have to be positioned side by side?

What I tried: Arrangements of $n$ distinct objects in a circle as $(n-1)!$

So arrangement of total $12$ men and women and children in a circle is $(12-1)!=11!$

Is my solution right? If not, then how do I solve it?

The given answer is $4!3!5!$.

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  • $\begingroup$ by positioning side by side does it mean no people of one type are together $\endgroup$
    – Albus Dumbledore
    Commented Oct 3, 2020 at 3:58
  • $\begingroup$ Still i have a confusion on that line. Whether one type sit together or not sit together. $\endgroup$
    – jacky
    Commented Oct 3, 2020 at 4:12
  • $\begingroup$ your answer is wrong in either case $\endgroup$
    – Albus Dumbledore
    Commented Oct 3, 2020 at 4:22
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    $\begingroup$ Your answer is correct if all refers to all the people. However, it is not clear why the question refers to $4$ men, $3$ women, and $5$ children. If the problem required that all the men, all the women, and/or all the children had to be positioned side by side, it would be more interesting. $\endgroup$
    – N. F. Taussig
    Commented Oct 3, 2020 at 8:36
  • $\begingroup$ Thanks N.F.Taussing . You are right because answer giben as $4!\times 3!\times 5!$. Can you please explain me. Thanks $\endgroup$
    – jacky
    Commented Oct 4, 2020 at 7:41

1 Answer 1

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In how many ways can $4$ men, $3$ women, and $5$ children be arranged in a circle if all have to be positioned side by side?

Given the wording of the question, there are indeed $(12 - 1)! = 11!$ distinguishable arrangements of the $4 + 3 + 5 = 12$ people up to rotation.

However, you said the stated answer is $4!3!5!$. This suggests that the author had a different question in mind, although the stated answer is not correct.

In how many ways can $4$ men, $3$ women, and $5$ children be arranged in a circle if all the men are seated together, all the women are seated together, and all the children are seated together?

We have three blocks of people to arrange in a circle. There are $(3 - 1)! = 2!$ distinguishable ways of arranging the blocks since either the blocks of men, women, and children are arranged in a clockwise order or anti-clockwise (counterclockwise) order. The four men can be arranged within their block in $4!$ ways, the three women can be arranged within their block in $3!$ ways, and the five children can be arranged within their block in $5!$ ways as we proceed clockwise around the circle. Hence, the number of distinguishable circular seating arrangements in which all the men sit together, all the women sit together, and all the children sit together is $2!5!4!3!$.

To obtain the stated answer, we would have to specify that the blocks of men, women, and children must be arranged in one of the two possible orientations of the blocks.

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