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In the figure above, P, Q, R, S and T are five different points no

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Bunuel
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In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   17 Aug 2017, 23:46
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In the figure above, P, Q, R, S and T are five different points no three of which are on the same straight line. How many different triangles can be drawn that have three of the five points as vertices if P must be one vertex of each triangle?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

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Luckisnoexcuse
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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   18 Aug 2017, 01:09
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Bunuel wrote:

In the figure above, P, Q, R, S and T are five different points no three of which are on the same straight line. How many different triangles can be drawn that have three of the five points as vertices if P must be one vertex of each triangle?

(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

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Attachment:
2017-08-18_1027_003.png


If P has to be one vertex then we are left with 4 points (Q,R,S,T) and we need to select 2 out of these 4 points to form a triangle with P as one of vertex
4C2 = 6

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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   18 Aug 2017, 02:18
Ans : d - 6

If P is one vertex , Then the remaining no of vertex are 4.

No of ways of selecting 2 points out of 4 = 4C2 =6.
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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   18 Aug 2017, 02:30
The solution should be Option D.

Can be solved using permutations and combinations formula to choose 2 more vertices from the remaining 4 points.

Can be written as

4C2 = 4!/2!(4-2)!
=4!/2!*2!
which comes to 6.
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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   18 Aug 2017, 02:48
Since one of the point P is fixed we need 2 points from the remaining 4 points

So 4C2 = 6

Hence answer = D
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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink] New post   03 Sep 2017, 05:35
The total triangles possible : 5C3
number of triangles with P as vertex = (total triangle possible) - (No. of triangle without P as vertex)
= 5C3 - 4C3
= 10 - 4
= 6
Option D is answer.
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Re: In the figure above, P, Q, R, S and T are five different points no [#permalink]
03 Sep 2017, 05:35
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