So now, divide out the redundancy. The whole variety of triangles created by these six traces is (6×5×four)/6, or 20. That’s the reply.
Right here’s the place math turns into highly effective. The identical process works for any variety of traces. What number of triangles are created by seven nonparallel traces? That’s (7×6×5)/6, or 35. What about 23 traces? (23×22×21)/6, or 1,771. How about 2,300 traces? That’s (2300×2299×2298)/6, which is a huge quantity: 2,025,189,100.
The identical calculation applies regardless of what number of traces there are. Examine that method to brute-force counting, which isn’t solely laborious and error-prone however supplies no option to test the reply. Math produces the answer and the rationale for it.
It additionally reveals that different issues are, at coronary heart, equivalent. Put balls of six totally different colours right into a bag. Pull out three. What number of totally different potential shade combos are there? 20, after all.
That’s combinatorics, and its helpful for fixing issues of this kind. It comes with its personal notation, to simplify the method of calculating, and entails loads of exclamation factors. The expression n! — “n factorial,” when mentioned aloud — describes the product of multiplying all of the integers from 1 to n. So 1! equals 1; 2! equals 2×1, or 2; three! equals three×2×1, or 6. And so forth.
In the issue by Dr. Loh, the calculation for the variety of triangles might be rewritten like this: 6!/(three!three!).
It may be written as C(6,three), which is learn as “6 select three.” Extra broadly, it’s mathspeak for the variety of methods to decide on three gadgets out of 6. It’s generalized into this type:
C(n, r) = n!/((n–r)!r!)
That’s the equation that college students memorize, the helpful shortcut. Give it a better look. The primary half — n!/(n–r)! — is what captures the 6×5×four within the triangle calculation. The r! is what eliminates the redundancies.