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by maihuna » Tue Mar 15, 2011 12:10 pm
The …ve sides of a pentagon have lengths of 2, 3, 4, 5, and 6
inches. Two pentagons are considered di¤erent only when the
positions of the side lengths are di¤erent relative to each other.
What is the total number of di¤erent possible pentagons that
could be drawn using these …ve side lengths?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120
Charged up again to beat the beast :)
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by anshumishra » Tue Mar 15, 2011 1:49 pm
maihuna wrote:The …ve sides of a pentagon have lengths of 2, 3, 4, 5, and 6
inches. Two pentagons are considered di¤erent only when the
positions of the side lengths are di¤erent relative to each other.
What is the total number of di¤erent possible pentagons that
could be drawn using these …ve side lengths?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120
It is like a ring. So, required no. = (n-1)!/2 = (5-1)!/2 = 12 B
Please note that a ring is different from a circular table scenario in the sense that two arrangements for example 1-2-3-4 is same as 1-4-3-2(if you flip the ring) .[/spoiler]
Thanks
Anshu

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by force5 » Tue Mar 15, 2011 2:11 pm
one more for 12....same reason
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