Simulation · Maths · Class 11

Patterns in the table of nCr

From the lesson Properties of combinations in Permutations and Combinations. Change the values and watch what happens.

The idea behind it

NCERT §6.4

nCr = nC(n − r): choosing r objects to take is the same as choosing n − r objects to leave. So 10C8 = 10C2 = 45.
If nCa = nCb, then a = b or a + b = n. If nC7 = nC5, then n = 12, and 12C2 = 66.
Pascal's rule: nCr + nC(r − 1) = (n + 1)Cr. For example 5C2 + 5C1 = 10 + 5 = 15 = 6C2.
The values nC0, nC1, …, nCn rise to the middle and fall back symmetrically. For n = 8 they are 1, 8, 28, 56, 70, 56, 28, 8, 1, with the largest, 70, at r = 4.
Taking at least one of 6 different things: 6C1 + 6C2 + … + 6C6 = 6 + 15 + 20 + 15 + 6 + 1 = 63.

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