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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.

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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 02 Aug 2015, 09:03
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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman

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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 02 Aug 2015, 11:03
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honchos wrote:
The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman


CONCEPT: The number of ways to arrange n elements around a circular Manner is (n-1)!.

But the number of ways to arrange n elements in circular manner where the arrangement can be flipped around (e.g. Necklace, Garland, Bracelet) = (n-1)!/2.


The pentagon in the problem above can be flipped over (Clockwise Arrangement of Sides = Counter-clockwise arrangement), the number of distinct ways to arrange the 5 sides = (5-1)!/2 = 12.

Answer: option B
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 30 Oct 2016, 08:34
Bunuel,

Can you please verify if the OA given is correct. In Jeff Sackman Extreme challenge PDF the answer given is 24. The OE is as follows:

Start by thinking of this as a typical permutations problem.
How many ways could the numbers 2, 3, 4, 5, and 6 be arranged in di¤erent
orders? The answer to that question would be 5!: the number permutations
of 5 di¤erent objects. However, since "two pentagons are considered di¤erent
only when the positions of the side lengths are di¤erent relative to each other,"
that eliminates some of those 5! permutations.
For instance, the following two arrangements of side lengths are the same:
2, 3, 4, 5, 6
6, 2, 3, 4, 5
In a typical permutations question, they would be counted as di¤erent. Here,
the side lengths are the same relative to each other. In fact, for every possible
pentagon, there are 5 "typical permutation" ways of counting it; in addition to
the above, there are:
5, 6, 2, 3, 4
4, 5, 6, 2, 3
3, 4, 5, 6, 2
Thus, 5! overstates the number of possible pentagons by a factor of 5, so the
total number of pentagons is:
5!
5 = 4! = 4  3  2  1 = 24, choice (C).
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 13 Nov 2016, 04:52
vrgmat wrote:
Bunuel,

Can you please verify if the OA given is correct. In Jeff Sackman Extreme challenge PDF the answer given is 24. The OE is as follows:

Start by thinking of this as a typical permutations problem.
How many ways could the numbers 2, 3, 4, 5, and 6 be arranged in di¤erent
orders? The answer to that question would be 5!: the number permutations
of 5 di¤erent objects. However, since "two pentagons are considered di¤erent
only when the positions of the side lengths are di¤erent relative to each other,"
that eliminates some of those 5! permutations.
For instance, the following two arrangements of side lengths are the same:
2, 3, 4, 5, 6
6, 2, 3, 4, 5
In a typical permutations question, they would be counted as di¤erent. Here,
the side lengths are the same relative to each other. In fact, for every possible
pentagon, there are 5 "typical permutation" ways of counting it; in addition to
the above, there are:
5, 6, 2, 3, 4
4, 5, 6, 2, 3
3, 4, 5, 6, 2
Thus, 5! overstates the number of possible pentagons by a factor of 5, so the
total number of pentagons is:
5!
5 = 4! = 4  3  2  1 = 24, choice (C).


Any updates about the topic? What is the right answer?

I could not understand the (n-1)! / 2 formula, since all circular arrangements can be flipped out? I appreciate if you can advice some problems about the issue?

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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 29 Nov 2016, 22:37
Hi Bunuel,

Please explain how the arrangements mentioned below are not different in this case..I think I am not interpreting the below mentioned line correctly-

"two pentagons are considered different only when the positions of the side lengths are different relative to each other,"
that eliminates some of those 5! permutations.
For instance, the following two arrangements of side lengths are the same:
2, 3, 4, 5, 6
6, 2, 3, 4, 5
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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 29 Nov 2016, 23:45
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honchos wrote:
The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman


Tricky one. :-D

5 sides of that pentagon are \(a,b,c,d,e\)

The totale numbers of possibles pentagons are \(5 \times 4 \times 3 \times 2 \times 1 =5!\)

Note that each possible selection of \((a,b,c,d,e)\), there are 4 other possible selections that are formed by arranged around each element. For example these 5 selections are considered the same pentagons
\(\begin{split}
2,3,4,5,6\\
6,2,3,4,5\\
5,6,2,3,4\\
4,5,6,2,3\\
3,4,5,6,2\\
\end{split}\)

Hence, in order to count the possible different selections, we need to divide the result by 5: \(5!/5=4!\)

Note that each pentagon could be fliped over, so both are considered the same. For example, these 2 selections are considered the same pentagons
\(\begin{split}
2,3,4,5,6\\
6,5,4,3,2\\
\end{split}\)

So, we also need to divide the result by 2 to avoid flipped over ones: \(4!/2=12\)

The final result are 12. The answer is B.

If the result are 24, this means that it covers the flipped over pentagons.
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 18 Jun 2017, 23:22
honchos wrote:
The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman


I am confident that the OA is wrong. The answer must be 24. Flipping over, or counting both clockwise and counter-clockwise arrangements the same is true only in case of necklaces, as they can be actually flipped over and worn. In this case of Drawing a pentagon on the paper, flipping over will result into nothing. When flipped over, you will get a plain paper with nothing. What I mean to say is that here the clockwise and counter-clockwise orientations definitely differ.
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 02 Jul 2017, 03:32
honchos wrote:
The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman


@chetanu: Hi, I got my answer as 24, but explanation given below involves the concept of flipping over hence dividing by 2, could you pls help how a pentagon drawn on paper could be flipped over.
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.  [#permalink]

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New post 13 Jul 2017, 16:23
ShashankDave wrote:
honchos wrote:
The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches. Two pentagons are considered different only when the positions of the side lengths are different relative to each others. What is the total number of different possible pentagons that could be drawn using these five side lengths ?
(A) 5
(B) 12
(C) 24
(D) 32
(E) 120

Source: Jeff Sackman


I am confident that the OA is wrong. The answer must be 24. Flipping over, or counting both clockwise and counter-clockwise arrangements the same is true only in case of necklaces, as they can be actually flipped over and worn. In this case of Drawing a pentagon on the paper, flipping over will result into nothing. When flipped over, you will get a plain paper with nothing. What I mean to say is that here the clockwise and counter-clockwise orientations definitely differ.


Actually, if you flip the pentagon, you get its reflection. To simplify things, try to imagine 2 equal triangles looking like the photo attached.

They might look different since one is the flipped version of the other, but they are equal, and should be counted as one. Same for the pentagons. If you flip them, basically you created the same thing from a different perspective, that doesn't change the fact that they are the same. Therefore, among these 24 results, 12 are just the flipped version and should not be counted.
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Re: The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.   [#permalink] 13 Jul 2017, 16:23
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The five sides of a pentagon have lengths of 2, 3, 4, 5 and 6 inches.

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