Counting Methods

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koby_gen: Junior | Next Rank: 30 Posts; Posts: 13; Joined: Fri Jan 07, 2011 10:14 pmFollowed by: 1 member; Thanked: 2 times

by koby_gen » Fri Jan 21, 2011 4:44 am

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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Source: Beat The GMAT — Problem Solving | Fri Jan 21, 2011 4:44 am

rohu27: Legendary Member; Posts: 586; Joined: Tue Jan 19, 2010 4:38 amFollowed by: 5 members; Thanked: 31 times; GMAT Score:730

by rohu27 » Fri Jan 21, 2011 5:44 am

Is OA A? Let me know so that I can post the explanation.
Dnt want to confuse you more with a wrong explanation if it isnt

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fskilnik@GMATH: GMAT Instructor; Posts: 1449; Joined: Sat Oct 09, 2010 2:16 pmFollowed by: 33 members; Thanked: 59 times

by fskilnik@GMATH » Fri Jan 21, 2011 6:04 am

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

Hi there!

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.

Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
English-speakers :: https://www.gmath.net
Portuguese-speakers :: https://www.gmath.com.br

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DarkKnight: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Mon Sep 15, 2008 5:54 pm; Thanked: 3 times

by DarkKnight » Fri Jan 21, 2011 11:31 am

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: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Fri Jan 21, 2011 12:20 pm

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 ways to arrange n elements around a circular table is (n-1)!.

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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tomada: Master | Next Rank: 500 Posts; Posts: 406; Joined: Mon Jan 25, 2010 11:36 amFollowed by: 4 members; Location: Syracuse, NY; Thanked: 23 times; GMAT Score:740

by tomada » Fri Jan 21, 2011 2:10 pm

Can someone please indicate the source of this question?

I'm really old, but I'll never be too old to become more educated.

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GMATGuruNY: GMAT Instructor; Posts: 15539; Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members; Location: New York, NY; Thanked: 13060 times; GMAT Score:790

by GMATGuruNY » Mon Jan 24, 2011 6:53 pm

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.

I was asked to explain further why the following is true:

(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: Senior | Next Rank: 100 Posts; Posts: 79; Joined: Mon Sep 15, 2008 5:54 pm; Thanked: 3 times

by DarkKnight » Mon Jan 24, 2011 7:48 pm

GMATGuruNY wrote:
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.
I was asked to explain further why the following is true:

(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?

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 already

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