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+n-2)
C) \(\frac{{mn(m+n-2)}}{2}\)
D) \(\frac{{(m+n)(m+n-1)(m+n-2)}}{6}\)
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(m-1)}/2 ways to select two of the m points (divide by 2 to strip out the double-counted 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(n-1)}/2 ways to select two of the n points. The total number of ways to select all three vertices of the triangle is thus:
\(\frac{m(m-1)}{2}*n+m*\frac{n(n-1)}{2}\)
\(\frac{mn(m-1)+mn(n-1)}{2}\)
\(\frac{mn(m-1+n-1)}{2}\)
\(\frac{mn(m+n-2)}{2}\)
The correct answer is (C).