The five sides of a pentagon have lenghts of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lenghts are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lenghts ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120
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fskilnik@GMATH
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Hi there!koby_gen wrote:The five sides of a pentagon have lenghts of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lenghts are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lenghts ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120
Without loss of generality we may assume that the side of length 6 is positioned (say "horizontally") and we are looking for the number of possible ordered choices (say clockwise chosen) for the lenghts of the other 4 sides (called A, B, C and D in my figure).
You have 4 choices for side A (2,3, 4 or 5); this side chosen, you have 3 choices for side B, then 2 choices for side C (and side D is left with the last length among the ones offered).
From the Multiplicative Principle, we have 4*3*2 = 24 possible choices.
The answer seems to be 24, but it is not. Please note from my second drawings that (according to the question stem) 24 is "duplicated" (I mean: every real distinct configuration was counted twice), therefore the answer is 12.
Regards,
Fabio.
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DarkKnight
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Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24.
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GMATGuruNY
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The number of ways to arrange n elements around a circular table is (n-1)!.DarkKnight wrote:Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24.
But the number of ways to arrange n elements around a ring is (n-1)!/2.
The ring gives us 1/2 the number of distinct arrangements because, unlike a table, a ring can be flipped over.
For example:
Given the 4 elements ABCD, there are (4-1)! = 3! = 6 ways to arrange them around a circular table.
Among these arrangements are ABCD and DBCA.
But around a ring, ABCD and DBCA represent the same arrangement, because when the ring is flipped over, ABCD becomes DBCA and DBCA becomes ABCD.
So around a ring, the number of distinct arrangements is cut in 1/2, as given in the formula above.
Assuming that -- like a ring -- the pentagon in the problem above can be flipped over, the number of distinct ways to arrange the 5 sides = (5-1)!/2 = 12.
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GMATGuruNY
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I was asked to explain further why the following is true:DarkKnight wrote:Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24.
(the number of distinct arrangements around a ring) = 1/2*(the number of distinct arrangements around a circular table)
Please see the picture below.
Let B = blue, G = green, P = purple, R = red.
Around a table, the 2 arrangements shown above -- BGPR and BRPG -- represent 2 distinct arrangements. These 2 distinct arrangements will be included when we count that there are (n-1)! = (4-1)! = 6 ways to arrange the four colors around the table.
Now imagine rotating the ring around the axis shown in the picture. When the ring is rotated 180 degrees, the arrangement on the left (BGPR) becomes the arrangement on the right (BRPG). Thus, around a ring, each placement of the four elements yields not one but two distinct arrangements. The result is that the number of distinct arrangements is cut in half. The number of ways to arrange the four colors around a ring = (n-1)!/2 = (4-1)!/2 = 3.
Does this help?
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DarkKnight
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Thats a neat concept Mitch. Your figures explained it all. Didn't really come across any problem revolving around a "ring" but will certainly keep in mind. Thank you much. Hope you are ready for some more snow in NY. I am dreading alreadyGMATGuruNY wrote:I was asked to explain further why the following is true:DarkKnight wrote:Why can't we apply the principle of arranging five people around a circular table?? According to that the answer should be (n-1)! = 4!= 24.
(the number of distinct arrangements around a ring) = 1/2*(the number of distinct arrangements around a circular table)
Please see the picture below.
Let B = blue, G = green, P = purple, R = red.
Around a table, the 2 arrangements shown above -- BGPR and BRPG -- represent 2 distinct arrangements. These 2 distinct arrangements will be included when we count that there are (n-1)! = (4-1)! = 6 ways to arrange the four colors around the table.
Now imagine rotating the ring around the axis shown in the picture. When the ring is rotated 180 degrees, the arrangement on the left (BGPR) becomes the arrangement on the right (BRPG). Thus, around a ring, each placement of the four elements yields not one but two distinct arrangements. The result is that the number of distinct arrangements is cut in half. The number of ways to arrange the four colors around a ring = (n-1)!/2 = (4-1)!/2 = 3.
Does this help?
.
