Let ABC be a triangle. Consider four points p_ 1 , p_ 2 , p_ 3 , p_ 4 on the side AB, five points p_ 5 , p_ 6 , p_ 7 , p_ 8 , p_ 9 on the side B C , and four points p_ 10 , p_ 11 , p_ 12 , p_ 13 on the side AC. None of

Question

Quantrex Academy · Accepted Answer

Correct answer: 660

Points: 4 on AB, 5 on BC, 4 on AC. For a valid pentagon, no 3 vertices can be collinear, so at most 2 points from each side.
Only valid distribution

(a, b, c)

with

a+b+c = 5

and each

\le 2

is permutations of

(1, 2, 2)

:

(1,2,2)

:

\binom{4}{1}\binom{5}{2}\binom{4}{2} = 4 \cdot 10 \cdot 6 = 240

(2,1,2)

:

\binom{4}{2}\binom{5}{1}\binom{4}{2} = 6 \cdot 5 \cdot 6 = 180

(2,2,1)

:

\binom{4}{2}\binom{5}{2}\binom{4}{1} = 6 \cdot 10 \cdot 4 = 240

Total

= 240 + 180 + 240 = 660

.

Let ABC be a triangle. Consider four points p_ 1 , p_ 2 , p_ 3 , p_...

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