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Re: A total of m different points are selected on a particular line, and a [#permalink]
Hi Lucky,

I am getting a slightly different answer.

N * mc2=

mc2= m* m-1 * (m-2) ! / 2! * (m-2)! => (mn) (m-1)/2----1

Same as above

(mn) (n-1)/2 -----------2


Adding 1 and 2

I am getting.

(mn) (m+n-2)/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+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.


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+n-2)
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A total of m different points are selected on a particular line, and a [#permalink]
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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*(n-1)*n + n*(m-1)*m)/2 = mn*(m+n-2)/2
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Re: A total of m different points are selected on a particular line, and a [#permalink]
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.
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Re: A total of m different points are selected on a particular line, and a [#permalink]
IMO C.

mC2*nC1+nC2*mC1.
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A total of m different points are selected on a particular line, and a [#permalink]
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.


A similar question can be found here https://www.quora.com/How-many-triangle ... ite-points

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+7-2) /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
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Re: A total of m different points are selected on a particular line, and a [#permalink]
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 am not able to understand this point and triangle concept of combinations..can somebody explain the basics of this?
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Re: A total of m different points are selected on a particular line, and a [#permalink]
There are 2 Scenarios. You have 2 Parallel Lines. Say M = No. of Different Points on the Top Parallel Line. And N = No. of Different Points on the Bottom Parallel Line.

Scenario 1: You must Choose 2 out of the M Different Points on the Top Parallel Line to create 1 Side of a Triangle. The Total No. of Different Combinations of ways to Choose 2 out of M Different Points will Give you all the different ways to create that 1 Side.

Then, you will have N Different Options for the 3rd Vertex on the Bottom Parallel Line.

"M Choose 2" * N Options


Scenario 2: You do the Same thing, but this time Choose 2 Points on the Bottom Parallel Line. Same Logic.

"N Choose 2" * M Options


Add the 2 Scenarios Together (its Either/Or)

(m! / (m - 2)! * 2!) * n + (n! / (n - 2)! * 2!) * m

Expand the Factorial! in the Numerator of both expressions so that you can Cancel out part of the Denominator.


(m * (m - 1) * (m - 2)! / (m - 2)! * 2! ) * n + (n * (n - 1) * (n - 2)! / (n - 2)! * 2!) * m


the (m-2)! in the 1st Expression and the (n-2)! in the 2nd Expression both cancel in the NUM and DEN

Since both Expressions have as the DEN: 2! = 2

you can ADD the 2 Fractions together

[ m * (m-1) * n + n * (n - 1) * m ] / 2

[ (m^2 - m) * n + (n^2 - n) * m ] / 2

[ m^2n - mn + mn^2 - mn ] / 2

PULLING OUT the Common Factor of mn

mn ( m + n - 2) / 2

Answer Choice C

A quicker way might be to Plug In Numbers and Test Answer Choices
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Re: A total of m different points are selected on a particular line, and a [#permalink]
Assume 3 points on one line, 3 also on the other.

3 points can be selected

6!/3!3! = 20 ways

Subtract sets of 3 points that lie on the same line.

Each line has 1 set of these to be subtracted, so

20-2= 18 triangles.

Substituting m=3 and n=3 into answer C

3*3*(3+3-2)/2 = 36/2=18

Answer C matches

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