Combinatorics and Newton’s binomial
Answers are for parents and teachers.
1
A four-digit PIN does not start with 0 and has all digits different. How many such PINs are there?
9 · 9 · 8 · 7 = 4536 (9 ways for the first digit, then 9, 8, 7).
2
Nine points are given in a plane, no three of them on one line. How many triangles have their vertices at these points?
84
3
Expand (x − 1)⁴.
x⁴ − 4x³ + 6x² − 4x + 1.
4
Find the constant term (the term without x) in the expansion of (x + 1/x)⁶.
Tₖ₊₁ = C(6, k)x⁶⁻ᵏ x⁻ᵏ = C(6, k)x⁶⁻²ᵏ; 6 − 2k = 0 gives k = 3, and C(6, 3) = 20.
5
Why is C(n, 0) + C(n, 1) + … + C(n, n) = 2ⁿ?
The left side adds the numbers of subsets with 0, 1, …, n elements of an n-element set, that is, the number of all subsets, which is 2ⁿ. Expanding (1 + 1)ⁿ with Newton’s binomial gives the same.