fskilnik wrote:

In how many ways four men, two women and one child can sit at a circular 7-seats table if the child is the only person to be seated between the two women?

(A) 24

(B) 36

(C) 48

(D) 96

(E) 240

Source:

http://www.GMATH.net\(?\,\,\,:\,\,\,\,\# \,\,{\text{circular}}\,\,{\text{permutations}}\,\,{\text{with}}\,\,{\text{restrictions}}\)

Alternate solution:

Let´s imagine a

linear version (=row), but "connecting the first seat to the last one" (so that after the last seat we have again the first one).

There are 7 seats in which the child could be seated.

Once (any) one of the 7 seats is chosen, there are 2 ways to seat the two women.

(If the child is in the 7th seat, W1 will be in the 6th, W2 in the 1st... or vice-versa!)

Once the child and the two women are seated, there are 4! ways of seating the men.

Using the

Multiplicative Principle, we have 7*2*4! ways of seating these people

in the linear version.

The "

linear to circular migration" is done dividing 7*2*4! by the number of objects to be circularized (7),

checking the "connection" created earlier do not give rise to unwanted configurations: it does not! (*)

Hence:

\(? = \frac{{7 \cdot 2 \cdot 4!}}{7} = 48\)

(*) Typical problem: when A and B cannot stay next to each other, in the linear version you cannot allow one of them to be in

the first place and the other in the last place, because when the connection is established they would violate the restriction!

This solution follows the notations and rationale taught in the GMATH method.

Regards,

Fabio.

_________________

Fabio Skilnik :: https://GMATH.net (Math for the GMAT) or GMATH.com.br (Portuguese version)

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