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In how many ways four men, two women and one child can sit at a circul

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In how many ways four men, two women and one child can sit at a circul  [#permalink]

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New post 24 Sep 2018, 13:26
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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

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Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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Re: In how many ways four men, two women and one child can sit at a circul  [#permalink]

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New post 24 Sep 2018, 13:54
Let women be numbered as w1 and w2 and child be as c
Arrangement can be done as w1cw2 or w2cw1 I.e. 2 ways

Now group the women children as one so in addition to 4 other men there are 5 entities to be arranged in a circle which can be done in (n-1)! Ways = (5-1)!= 4!= 24 ways

So total ways = 2*24 = 48 ways

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In how many ways four men, two women and one child can sit at a circul  [#permalink]

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New post 24 Sep 2018, 15:27
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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

\(?\,\,\,:\,\,\,\,\# \,\,{\text{circular}}\,\,{\text{permutations}}\,\,{\text{with}}\,\,{\text{restrictions}}\)

Let the child be placed in any seat.
Once this is done, there are 2 ways of placing the women (W1 to-the-right of the child, W2 to-the-left of the child... and vice-versa).
Once the child and the women are seated, there are 4! ways of placing the men.

From the Multiplicative Principle:

\({\text{?}}\,\,\, = \,\,\,2 \cdot 4!\,\,\, = \,\,48\)


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

Regards,
Fabio.
_________________

Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

GMATH Teacher
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Joined: 12 Oct 2010
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In how many ways four men, two women and one child can sit at a circul  [#permalink]

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New post 24 Sep 2018, 15:32
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).

Image

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 :: http://www.GMATH.net (Math for the GMAT)
Course release PROMO : finish our test drive till 31/Oct with (at least) 60 correct answers out of 92 (12-questions Mock included) to gain a 60% discount!

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