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A total of m different points are selected on a particular line, and a
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01 May 2015, 01:19
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61% (01:46) correct 39% (01:55) wrong based on 287 sessions
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A total of m different points are selected on a particular line, and a
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01 May 2015, 07:55
Bunuel wrote: A total of m different points are selected on a particular line, and a total of n different points are selected on another line parallel to the first, where each of m and n is greater than 1. In how many different ways can a triangle be made with its vertices at three of the selected points?
A) m^2n+mn^2
B) mn(m+n2)
C) \(\frac{{mn(m+n2)}}{2}\)
D) \(\frac{{(m+n)(m+n1)(m+n2)}}{6}\)
E) \(\frac{{(m+n)!}}{{m!n!}}\)
Kudos for a correct solution. case 1: One point on particular line and other 2 points on parallel line. #ways= m* nc2 Case 2: one point on parallel line and other 2 on particular line. n* mc2 Total m* nc2+ n* mc2 mn(m+n2) B
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Re: A total of m different points are selected on a particular line, and a
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01 May 2015, 09:47
Hi Lucky, I am getting a slightly different answer. N * mc2= mc2= m* m1 * (m2) ! / 2! * (m2)! => (mn) (m1)/21 Same as above (mn) (n1)/2 2 Adding 1 and 2 I am getting. (mn) (m+n2)/2 Please could you highlight me where am I going wrong. Lucky2783 wrote: Bunuel wrote: A total of m different points are selected on a particular line, and a total of n different points are selected on another line parallel to the first, where each of m and n is greater than 1. In how many different ways can a triangle be made with its vertices at three of the selected points?
A) m^2n+mn^2
B) mn(m+n2)
C) \(\frac{{mn(m+n2)}}{2}\)
D) \(\frac{{(m+n)(m+n1)(m+n2)}}{6}\)
E) \(\frac{{(m+n)!}}{{m!n!}}\)
Kudos for a correct solution. case 1: One point on particular line and other 2 points on parallel line. #ways= m* nc2 Case 2: one point on parallel line and other 2 on particular line. n* mc2 Total m* nc2+ n* mc2 mn(m+n2) B
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A total of m different points are selected on a particular line, and a
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01 May 2015, 11:21
we either choose 1 point on first line AND 2 points on second line or vice versa, resulting sum ends up being as follows: mC1*nC2 + nC1*mC2 = (m*(n1)*n + n*(m1)*m)/2 = mn*(m+n2)/2 C



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Re: A total of m different points are selected on a particular line, and a
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04 May 2015, 02:14
Bunuel wrote: A total of m different points are selected on a particular line, and a total of n different points are selected on another line parallel to the first, where each of m and n is greater than 1. In how many different ways can a triangle be made with its vertices at three of the selected points?
A) m^2n+mn^2
B) mn(m+n2)
C) \(\frac{{mn(m+n2)}}{2}\)
D) \(\frac{{(m+n)(m+n1)(m+n2)}}{6}\)
E) \(\frac{{(m+n)!}}{{m!n!}}\)
Kudos for a correct solution. MANHATTAN GMAT OFFICIAL SOLUTION:Smart Numbers SolutionThe values of m and n are undetermined in the problem, so select values for those variables, say m = 3 and n = 2. Let the three m points be designated A, B, and C, and the two n points X and Y. Using these variables, there are nine ways to construct the desired triangle: 2 points from the m set and 1 point from the n set: ABX, ABY, ACX, ACY, BCX, BCY 2 points from the n set and 1 point from the m set: AXY, BXY, CXY The desired answer is thus 9. Check the answer choices by plugging in m = 3 and n = 2: (A) (9)(2)+(3)(4)=18+ too big (B) (3)(2)(3)= too big (C) (3)(2)(3)/2=9 (D) (5)(4)(3)/6=10 (E) 5!/(3!2!)=(5*4)/(2*1)=10 Only choice (C) yields the desired answer. Algebraic/Geometric SolutionThis can be done algebraically… although the math is so annoying that we don’t recommend it! The triangle can be formed in one of two principal ways: (i) by selecting two of the m points and one of the n points, or (ii) by selecting one of the m points and two of the n points. (Three points from the same line cannot be selected; doing so would create a line segment rather than a triangle.) In case (i), there are {m(m1)}/2 ways to select two of the m points (divide by 2 to strip out the doublecounted possibilities of having, say, vertices at (3, 0) and (5,0) or vertices at (5,0) and (3,0)—these are the same option). There are n ways to select one of the n points. In case (ii), there are m ways to select one of the m points, and {n(n1)}/2 ways to select two of the n points. The total number of ways to select all three vertices of the triangle is thus: \(\frac{m(m1)}{2}*n+m*\frac{n(n1)}{2}\) \(\frac{mn(m1)+mn(n1)}{2}\) \(\frac{mn(m1+n1)}{2}\) \(\frac{mn(m+n2)}{2}\) The correct answer is (C).
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Re: A total of m different points are selected on a particular line, and a
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12 Oct 2016, 18:04
Bunuel wrote: A total of m different points are selected on a particular line, and a total of n different points are selected on another line parallel to the first, where each of m and n is greater than 1. In how many different ways can a triangle be made with its vertices at three of the selected points?
A) m^2n+mn^2
B) mn(m+n2)
C) \(\frac{{mn(m+n2)}}{2}\)
D) \(\frac{{(m+n)(m+n1)(m+n2)}}{6}\)
E) \(\frac{{(m+n)!}}{{m!n!}}\)
Kudos for a correct solution. I don't get it..is A  (m^2)*n+m*(n^2) or is it (m^2n)...??? anyways...i think best strategy here is to give m and n some values, while considering that m and n must be > than 1. suppose m=3 and n=2. in this case, we can select 1 point from m line, and 2 points from n line> 3C1 * 2C2 = 3*1 or we can select 2 points from m line, and 1 point from n line > 3C2 * 2C1 > 3*2 = 6 total, 3+6=9 possible triangles. A  doesn't matter what is the correct form of the answer, but it yields a way too bigger number...out. B  3*2(3+22) = 6*3 = 18. nope. C  basically B but divided by 2. 9  might be the answer. D  (3+2)(3+21)(3+22) = 5*4*3  way too big. E  (m+n)! = 5!. 5!/3!2! = 5*4*3!/1*2*3! => 5*2 = 10  not good. luckily, C works out. C must be the answer. p.s. i tried first by putting: mC2 * nC1 + mC1*nC2 > nC1 and mC1 is n and n so definitely m multiplied by smth + n multiplied by smth...since it might have gotten into wild territories, I decided to take another approach, the one that i described above.



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Re: A total of m different points are selected on a particular line, and a
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30 Dec 2016, 09:07
IMO C.
mC2*nC1+nC2*mC1.



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A total of m different points are selected on a particular line, and a
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16 Jun 2017, 02:55
Bunuel wrote: A total of m different points are selected on a particular line, and a total of n different points are selected on another line parallel to the first, where each of m and n is greater than 1. In how many different ways can a triangle be made with its vertices at three of the selected points?
A) m^2n+mn^2
B) mn(m+n2)
C) \(\frac{{mn(m+n2)}}{2}\)
D) \(\frac{{(m+n)(m+n1)(m+n2)}}{6}\)
E) \(\frac{{(m+n)!}}{{m!n!}}\)
Kudos for a correct solution. A similar question can be found here https://www.quora.com/Howmanytriangle ... itepoints The question basically asks you to apply combinatorics so if we have m points on our first line and n points on our second line then we choose 1 point from the m points on the first line and m 1 point from n points and then choose x points from the remainder of M + N  2 Mc1 * Nc1 * M +N 2 / 2 So if we had 8 points on m and 7 points on n then our formula for the number of distinct triangles would be 8c1 * 7c1 * (8+72) /2! ( we use this denominator because order does not matter *see diagram* so even if we labeled each distinct point on each line it wouldn't matter what order you pick it in because it forms the same triangle). Here is a counter example, if the GMAT asks you how many ways could you arrange Freddy, Shaggy, Daphne, Scooby and Velma well then order does matter because the position of each member makes a distinct arrangement. Freddy, Shaggy, Daphne, Scooby and Velma is NOT the same as Shaggy, Freddy, Daphne, Scooby and Velma. Thus C



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Re: A total of m different points are selected on a particular line, and a
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28 Oct 2018, 13:50
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