A decision tree that grows one stage at a time: every choice multiplies the leaves. Then the addition rule for either-or cases.

Fill k slots from n items and watch the choices shrink — n, n − 1, n − 2 — or stay at n when repeats are allowed.

List every ordered pick, then gather the k! orderings of each set into one row. What is left is nCr.

Build each row by adding the two numbers above. Find C(n, k), its mirror image, the hockey stick and the row sums of 2ⁿ.

Expand (a + b)ⁿ by picking a or b from each factor, and see why the coefficients are C(n, k) — with (a + b)³ as a cube cut into 8 blocks.

Share n identical items among k bins by dragging k − 1 bars through a row of stars. Every arrangement is one distribution.

Count a union of overlapping sets by adding, subtracting and adding back — with a tally on every Venn region so nothing is counted twice.

More pigeons than holes forces a collision. Then a surprise: in any group, two people have shaken the same number of hands.

The code side of permutations: build every ordering with recursion and backtracking, one choice per level of the call tree.

Include or exclude each element in turn: the 2ⁿ leaves of the recursion tree are exactly the subsets that Pascal's rows count.

n! = n × (n − 1)! computed by recursion, with the call stack growing and unwinding — the number every page here divides by.