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Fifteen dots are evenly spaced on the circumference of a circle. How

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Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 06 Mar 2015, 06:34
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A
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C
D
E

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Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 07 Mar 2015, 19:46
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Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


Kudos for a correct solution.


hi all,
there are two ways we can do this...

1) estimation/elimination process...


total ways=15C3=15!/12!3!=455...
so if total ways are 455 we eliminate any choice which is above this... so C,D and E are out...
pure logic - equilateral triangles are going to be way less than total triangles possible.... and choice A gives % of equilateral triangle almost 65% of total triangles...
so A can be eliminated ..only B left

2) pure mathematical way...


mark these points 1 to 15...
if all 15 points are equidistant, for an equilateral triangle the three points should be equidistant from each other and that will be possible only in one scenario.. when the dots are 5 portion away from each other...
so dots will be 1,6,11 or 2,7,12... and so on ..
15 points will give us only 5 sets of these values, as rest 10 will be repetition of these 5...
let me write down the five possible way..
1) 1,6,11
2) 2,7,12
3) 3,8,13
4) 4,9,14
5) 5,10,15
6) 6,11,1.. this is same as (1)..

ans 5 ways
lef ways =455-5=450 ans B....
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 08 Mar 2015, 14:40
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5
Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


Kudos for a correct solution.


MAGOOSH OFFICIAL SOLUTION:

Well, first of all, ignoring the type of triangle formed, how many combinations total? The easiest way to think about this is to use the Fundamental Counting Principle. For the first dot, 15 choices, then 14 left for the second choice, then 13 left for the third choice: that’s 15*14*13. But, that will count repeats: the same three dots could be chosen in any of their 3! = 6 orders, so we have to divide that number by 6. (NOTICE the non-calculator math here).
(15*14*13)/6

Cancel the factor of 3 in 15 and 6
(5*14*13)/2

Cancel the factor of 2 in the 14 and 2
(5*7*13) = 5*91 = 455

That’s how many total triangles we could create.

Of these, how many are equilateral triangles? Well, the only equilateral triangles would be three points equally spaced across the whole circle. Suppose the points are numbers from 1 to 15. From point 1 to point 6 is one-third of the circle — again, from point 6 to point 11, and from point 11 back to point 1. That’s one equilateral triangle. We could make an equilateral triangle using points
{1, 6, 11}
{2, 7, 12}
{3, 8, 13}
(4, 9, 14)
{5, 10, 15}

After that, we would start to repeat. There are five possible equilateral triangles, so 455 – 5 = 450 of these triangles are not equilateral.

Answer = (B)
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 22 Jan 2016, 01:32
Hi Bunuel,
Please, can you kindly help me to understand why the answer is B and not D. When I got 455 (i.e., (15*14*13/3*2*1)), I read it (i.e. 455) to be the number of slots for the combinations of three-dot equilateral triangles. So I went further to multiply the 455 by 3 to get 1365 from which I deducted 5 (i.e., the 5 possible equilateral triangles) to obtain 1360, answer (D).

Thank you

Solomon
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 22 Jan 2016, 02:00
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duahsolo wrote:
Hi Bunuel,
Please, can you kindly help me to understand why the answer is B and not D. When I got 455 (i.e., (15*14*13/3*2*1)), I read it (i.e. 455) to be the number of slots for the combinations of three-dot equilateral triangles. So I went further to multiply the 455 by 3 to get 1365 from which I deducted 5 (i.e., the 5 possible equilateral triangles) to obtain 1360, answer (D).

Thank you

Solomon


Hi,
any combination of three points will give you only one triangle..
if 1,2,3 are three such points, 123, 231,312 all are same triangle..
therefore when you have got 455 ways of choosing 3 triangle, these will give you 455 triangles as each way will give you exactly one unique triangle..

Hope it helped
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 24 Apr 2016, 22:52
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Trying to visualize it I solved it this way:

Equilateral triangle is:

XOOOOXOOOOXOOOO

So we have 15 elements with 12 and 3 repeating:
Hence C(3 and 12 identical out of 15) = C(3,15)=15!/(12!*3!)=455

No the only time we can get the equilateral triangle is as I showed earlier:

XOOOOXOOOOXOOOO
Looking at one side only:

XOOOO

There are only 5 ways the items can be arranged (i.e. XOOOO, OXOOO, OOXOO, OOOXO, OOOOX)

So the total number of triangles should be reduced by this number
and answer becomes 455-5=450.
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 27 Aug 2018, 11:06
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Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


The total number of triangles is equal to C(15,3) = 455.

Exactly 5 of them are equilateral triangles. The image attached shows the reason.

? = 450

The above follows the notations and rationale taught in the GMATH method.
Attachments

27Ago18_7r.gif
27Ago18_7r.gif [ 27.16 KiB | Viewed 5386 times ]


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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 29 Nov 2018, 08:56
Top Contributor
Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


KEY CONCEPT
If we connect ANY 3 dots, we'll get a unique triangle.
Since the order in which we select the dots does not matter, we can use COMBINATIONS.
We can select 3 dots from 15 dots in 15C3 ways
15C3 = (15)(14)(12)/(3)(2)(1) = 455

Now SOME of these 455 triangles will be equilateral triangles, so we must subtract from 455 the number of those triangles that are equilateral triangles.

IMPORTANT: Since the correct answer must be less than 455, we can ELIMINATE answer choices C, D and E

At this point, we COULD determine the number of equilateral triangles that are included among the 455 triangles we've counted (the above posters have already done so).
However, we could use the answer choices to our advantage.

Answer choice A (160) suggests that there are 295 equilateral triangles among the 455 triangles we've counted (since 455 - 295 = 160)
Answer choice B (450) suggests that there are 5 equilateral triangles among the 455 triangles we've counted (since 455 - 5 = 450)

If answer A is correct, then more than half of the 455 triangles are equilateral triangles. This doesn't seem right since MOST selections of 3 points will NOT yield an equilateral triangle.
So, ELIMINATE A

Answer: B

Cheers,
Brent
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 04 Nov 2019, 04:12
1
Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


Good = Total - Bad

Total:
From 15 points, the number of ways to choose 3 = 15C3 = (15*14*13)/(3*2*1) = 5*7*13.

Bad:
Here, a bad combination can be used to form an equilateral triangle.
Let the 15 points on the circumference be divided into 3 equally spaced groups:
A-B-C-D-E
F-G-H-I-J
K-L-M-N-O
Equilateral triangles can be formed as follows:
A-F-K (connecting the first point in each group)
B-G-L (connecting the second point in each group)
C-H-M (connecting the third point in each group)
D-I-N (connecting the fourth point in each group)
E-J-O (connecting the last point in each group)
5 ways

Good:
Total - Bad = (5*7*13) - 5 = 5(7*13 - 1) = 5*90 = 450

.
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How  [#permalink]

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New post 12 Nov 2019, 19:09
Bunuel wrote:
Fifteen dots are evenly spaced on the circumference of a circle. How many combinations of three dots can we pick from these 15 that do not form an equilateral triangle?

A. 160
B. 450
C. 910
D. 1360
E. 2640


Kudos for a correct solution.


The total number of triangles that can be formed (regardless of whether they are equilateral) is 15C3 = (15 x 14 x 13)/(3 x 2) = 5 x 7 x 13 = 455. Now let’s determine the number of triangles formed that are equilateral triangles.

Let A, B, C, D, E, F, G, H, I, J, K, L, M, N, and O be the 15 points that are evenly spaced on the circumference of the circle. Every pair of consecutive points (for example, A and B, G and H, etc.) form a 360/15 = 24-degree arc. In order for 3 points to form an equilateral triangle, they must evenly spaced among themselves on the circumference of the circle also. That is, each pair of the three points have to be 360/3 = 120 degrees apart. Since 120/24 = 5, the points should be 5 spaces apart from one another. For example, if A is one of the vertices of the equilateral triangle, then the next one should be F and the last one should be K. In other words, triangle AFK is an equilateral triangle. Using the same analogy, triangles BGL, CHM, DIN, and EJO are equilateral triangles and only these 5 (including triangle AFK) are equilateral triangles.

Since 455 triangles can be formed and 5 of these are equilateral, then the number of triangles that are not equilateral is 455 - 5 = 450.

Answer: B
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Re: Fifteen dots are evenly spaced on the circumference of a circle. How   [#permalink] 12 Nov 2019, 19:09
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