The binomial distribution, a few chapters ahead, leans on one number computed over and over: \binom{n}{k}, read "n choose k." Pascal's triangle is the fastest way to see where that number comes from. It also shows why the number never needs to be looked up from scratch.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Discrete Mathematics I — Logic, Proof, and Structure
Compute the binomial coefficient C(6,2).
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Why
C(6,2) = 6!/(2! times 4!) = 15.
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From Discrete Mathematics I — Logic, Proof, and Structure
Given C(6,2)=15 and C(6,3)=20, use Pascal's identity to find C(7,3).
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Why
Pascal's identity: C(7,3) = C(6,2) + C(6,3) = 15 + 20 = 35.
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Picture a club deciding who sits on a small committee. Pascal's triangle answers the same question over and over. Out of a group of a certain size, how many different committees of a certain size can be formed? Now suppose one new member joins the club, and the count needs updating without starting over from scratch. Every possible committee either includes the new member or leaves them out, never both, so the two cases can be counted separately and added. Include the new member, and the rest of the committee is chosen from the old members alone. Leave the new member out, and the whole committee is chosen from the old members alone. Add those two counts together and the update is done. That is the whole rule behind Pascal's triangle. Each entry is built from the two entries above it, because it is counting exactly those two cases.
Row n of Pascal's triangle lists \binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n} in order. Every interior entry is the sum of the two entries diagonally above it: \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}. This is not a coincidence of arithmetic. It reflects a counting argument: either the last of the n items is in the chosen group of k, or it isn't. That split divides every size-k group from n items into exactly those two disjoint cases.
Click any entry
Click an entry to see its value, its committee meaning, and the two parents above it that it's built from.
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Worked example
Reading an entry two ways
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1
Find \binom{5}{2} using the factorial formula, then check it against Pascal's triangle.
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2
\binom{5}{2} = \dfrac{5!}{2!\,3!} = \dfrac{120}{2 \times 6} = 10.
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3
Row 5, position 2 of Pascal's triangle reads 10. The two numbers agree, as they always will.
\boxed{\binom{5}{2} = 10}
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Worked example
Using the recurrence directly
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1
Given \binom{6}{2} = 15 and \binom{6}{3} = 20, find \binom{7}{3} without a fresh factorial computation.
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2
\binom{7}{3} is built from the two entries diagonally above it in row 6: \binom{6}{2} and \binom{6}{3}.
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3\boxed{\binom{7}{3} = \binom{6}{2} + \binom{6}{3} = 15 + 20 = 35}
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Where this goes next
Chapter 5 (The named discrete distributions)
The binomial pmf, P(X{=}k) = \binom{n}{k}p^k(1-p)^{n-k}, uses \binom{n}{k} to count exactly one thing: how many of the 2^n possible sequences of successes and failures have precisely k successes. Every entry of Pascal's triangle you read here is a rehearsal for reading that formula without flinching.
Check your understanding (counts toward your progress)
What is \binom{9}{2}?
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Why
\binom{9}{2} = \dfrac{9!}{2!\,(9-2)!} = 36.
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Explain the committee argument for why \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}.
AI grading isn't configured on this deployment.
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First mistake
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Why
Fix attention on one specific person among the n. Either that person is on the k-person committee (leaving k-1 seats to fill from the other n-1 people) or they aren't (leaving all k seats to fill from the other n-1) — every committee falls into exactly one of these two disjoint cases, so the counts add.
Fix your attention on one specific person and ask: are they on the committee, or not?
What is \binom{10}{0}?
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Why
\binom{10}{0} = \dfrac{10!}{0!\,(10-0)!} = 1.
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What is \binom{6}{2}?
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Why
\binom{6}{2} = \frac{6!}{2!4!} = 15.
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Practice: Computing and reasoning with binomial coefficients (5 problems · doesn’t affect your progress)
Pascal's triangle's row 4 is 1, 4, 6, 4, 1. Which single fact explains why the two ends of every row are always 1?
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It's not an arbitrary convention — \binom{n}{0} and \binom{n}{n} genuinely both equal 1 by the choosing definition.
It's not a coincidence — it follows directly from what \binom{n}{0} and \binom{n}{n} mean.
\binom{n}{1}=n in general, not 1 (except when n=1) — that's not why the ends are 1.
Why
Choosing 0 items (do nothing) or all n items (take everything) each has exactly one way, so \binom{n}{0}=\binom{n}{n}=1 always.
\binom{n}{0}=\binom{n}{n}=1 — there's exactly one way to choose none or all of n items
Compute \binom{7}{2}.
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Why
\binom{7}{2} = \frac{7\times6}{2\times1} = 21.
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Find the value of \binom{9}{6}.
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Why
By symmetry, \binom{9}{6}=\binom{9}{3}=\frac{9\times8\times7}{3\times2\times1}=84.
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Use the symmetry property \binom{n}{k}=\binom{n}{n-k} to switch to a smaller, easier computation.
Given \binom{6}{2}=15 and \binom{6}{3}=20, use the Pascal recurrence to find \binom{7}{3}.
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Why
By the recurrence, \binom{7}{3}=\binom{6}{2}+\binom{6}{3}=15+20=35.
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Why
\binom{7}{3}=\binom{6}{2}+\binom{6}{3} — add the two given parent values.
Without expanding factorials directly, use the symmetry property to rewrite \binom{12}{9} as \binom{12}{k} for a smaller k, then evaluate.
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Why
By symmetry, \binom{12}{9}=\binom{12}{3}=\dfrac{12\times11\times10}{6}=220 — far easier than expanding 9 terms directly.
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Why
\binom{12}{9}=\binom{12}{12-9}=\binom{12}{3} — far fewer terms to multiply.
\binom{12}{3}=\dfrac{12\times11\times10}{3\times2\times1}.