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.
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+2-2) = 6*3 = 18. nope.
C - basically B but divided by 2. 9 - might be the answer.
D - (3+2)(3+2-1)(3+2-2) = 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.