Binomial Coefficients and Pascal's Triangle

Counting "how many ways can I pick 3 stocks out of 20 for a basket" seems like it should require listing possibilities, but that becomes hopeless fast — even 20 choose 3 already has 1,140 combinations. Binomial coefficients give a closed formula for exactly this kind of count, and Pascal's triangle gives an almost magical way to build every one of these numbers just by repeated addition, without ever computing a single factorial.

An analogy: choosing a starting five, not a batting order

If you're choosing which 5 of 12 players will start a game — a plain selection where order doesn't matter, since "start" is start regardless of who's announced first — that's fundamentally a different, smaller count than choosing an ordered lineup of those 5 (where batting order matters). Binomial coefficients count the first kind: unordered selections. $\binom{n}{k}$ ("n choose k") is the number of distinct subsets of size $k$ you can pick from $n$ items, with no regard to the order you picked them in.

The formula, one symbol at a time

$$\binom{n}{k} = \frac{n!}{k!\,(n-k)!} ,$$

where $n!$ ("n factorial") means $n\times(n-1)\times\cdots\times1$, and $\binom{n}{k}$ reads "n choose k." In plain English: count all $n!$ orderings of the full group, then divide by $k!$ (to stop caring about the order within your chosen subset of size $k$) and by $(n-k)!$ (to stop caring about the order of the leftover, unchosen items). Pascal's triangle arranges these numbers in rows, row $n$ holding $\binom{n}{0}, \binom{n}{1}, \dots, \binom{n}{n}$, and each entry equals the sum of the two entries diagonally above it in the previous row:

$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} .$$

In plain English: to select $k$ items from $n$, either a specific item is in your selection (then you still need to pick the remaining $k-1$ from the other $n-1$) or it's not (then you need all $k$ from the other $n-1$) — these two cases are mutually exclusive and cover everything, so their counts simply add.

Worked example 1: choosing 3 stocks from 6 by hand

$\binom{6}{3} = \frac{6!}{3!\,3!} = \frac{720}{6\times6} = \frac{720}{36} = 20$. Cross-checking with the addition rule: $\binom{6}{3} = \binom{5}{2}+\binom{5}{3}$. Computing each: $\binom{5}{2}=\frac{120}{2\times6}=10$, $\binom{5}{3}=\frac{120}{6\times2}=10$. Sum: $10+10=20$ — matching exactly, confirming the recursive addition rule gives the same answer as the direct factorial formula, just built up from smaller cases.

Worked example 2: building three rows of Pascal's triangle

Row 0: $\binom{0}{0}=1$. Row 1: $\binom{1}{0}=1, \binom{1}{1}=1$. Row 2: $\binom{2}{0}=1$, $\binom{2}{1}=\binom{1}{0}+\binom{1}{1}=1+1=2$, $\binom{2}{2}=1$. Row 3: $\binom{3}{0}=1$, $\binom{3}{1}=\binom{2}{0}+\binom{2}{1}=1+2=3$, $\binom{3}{2}=\binom{2}{1}+\binom{2}{2}=2+1=3$, $\binom{3}{3}=1$. So row 3 reads 1, 3, 3, 1 — exactly the coefficients you'd get expanding $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$, which is the "binomial" in binomial coefficient: $\binom{n}{k}$ is literally the coefficient of $a^{n-k}b^k$ in the expansion of $(a+b)^n$.

Picture Pascal's triangle as a grid where each cell sums the two above it: the outer edges are always 1, and the interior numbers grow rapidly toward the middle of each row, mirroring the classic bell-curve shape that emerges from binomial probabilities as $n$ grows.

What this means in practice

Binomial coefficients are the raw counting machinery behind the binomial probability distribution (probability of exactly $k$ successes in $n$ trials involves $\binom{n}{k}$ directly), option-pricing binomial trees (the number of distinct up/down paths reaching a given node), portfolio construction from a universe ("how many distinct 30-stock portfolios from a 500-stock universe"), and virtually every probability brainteaser involving counting subsets. Fast, mental estimation of $\binom{n}{k}$ for small $k$ (like $\binom{n}{2}=n(n-1)/2$) is a standard interview skill.