In short

Each coefficient \binom{n}{k} in the expansion of (a+b)^n is a count: it counts the number of ways to pick k of the n binomial factors to contribute a "b" (with the remaining n-k factors contributing an "a"). Pascal's recurrence \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} is just a yes/no question about the last factor — either you picked its b (then choose k-1 more bs from the previous n-1) or you did not (then choose all k bs from the previous n-1). Two disjoint cases that exhaust everything, so they add. That is the entire reason Pascal's triangle works.

You already know that row n of Pascal's triangle gives the coefficients of (a+b)^n, and that the rule for building the triangle is "each entry is the sum of the two directly above it." But why does that addition rule produce the right numbers? Why is the coefficient of a^4 b^2 in (a+b)^6 exactly 15 and not 14 or 16? You can verify it by multiplying out, but verification is not understanding. The understanding comes from realising that those coefficients are not algebraic accidents — they are counts of something you can list on a piece of paper. Once you see what they count, Pascal's recurrence stops being a magical rule and becomes the only thing it could possibly be.

This article explains the counting picture. The companion article — Pascal's triangle: click a row, watch (a+b)ⁿ pop out — gives you the interactive widget. Here we go after the proof.

The combinatorial argument: each coefficient counts picks

Write (a+b)^n the long way:

(a+b)^n = \underbrace{(a+b)(a+b)(a+b)\cdots(a+b)}_{n \text{ factors}}

When you expand this product fully, you walk through the n brackets and from each one you must pick either the a or the b. Multiply the picks together; that is one term of the expansion. There are n independent yes/no choices, so there are 2^n such terms before you collect like ones.

Now sort the 2^n raw terms by how many bs appear in the pick. Suppose you chose the b from exactly k of the brackets and the a from the remaining n-k. The term you wrote down is

\underbrace{a \cdot a \cdots a}_{n-k \text{ copies}} \cdot \underbrace{b \cdot b \cdots b}_{k \text{ copies}} = a^{n-k}\, b^k

regardless of which particular k brackets you took the bs from. So every pick with exactly k bs collapses to the same monomial a^{n-k} b^k. The coefficient of a^{n-k} b^k in the expansion is therefore the number of ways to choose which k of the n brackets contribute a b.

That number has a name. It is "n choose k", written \binom{n}{k}.

Why is \binom{n}{k} the right count? By definition, \binom{n}{k} is the number of ways to pick a k-element subset from an n-element set, where order does not matter. The brackets are an n-element set. Picking which k of them give a b is exactly picking a k-element subset. So the coefficient is \binom{n}{k} — not by some derivation, but by pure matching of definitions.

That is the binomial theorem in one paragraph:

(a+b)^n = \sum_{k=0}^{n} \binom{n}{k}\, a^{n-k}\, b^k

The coefficients are not pulled from thin air or memorised from a table. They are counts of choices, and those choices live inside the product itself.

The eight picks for (a+b)(a+b)(a+b), grouped by how many b's are chosenThree boxes labelled (a+b) at the top, representing the three factors of (a+b) cubed. Below them, the eight possible ways to pick an a or a b from each box are listed in four groups: zero b's gives one way (aaa), one b gives three ways (baa, aba, aab), two b's gives three ways (bba, bab, abb), and three b's gives one way (bbb). The group sizes 1, 3, 3, 1 match row 3 of Pascal's triangle. $(a+b)(a+b)(a+b)$ — three factors, eight picks total a+b a+b a+b factor 1 factor 2 factor 3 0 b's — coefficient of $a^3$ a a a count = 1 = C(3,0) 1 b — coefficient of $a^2 b$ b a a a b a a a b count = 3 = C(3,1) 2 b's — coefficient of $a b^2$ b b a b a b a b b count = 3 = C(3,2) 3 b's — coefficient of $b^3$ b b b count = 1 = C(3,3) row 3 of Pascal's triangle: 1 3 3 1 $2^3 = 8$ raw picks, sorted into bins of size $\binom{3}{0}, \binom{3}{1}, \binom{3}{2}, \binom{3}{3}$
All $2^3 = 8$ picks for $(a+b)(a+b)(a+b)$, grouped by how many $b$s appear. Bins of size $1, 3, 3, 1$ — exactly row $3$ of Pascal's triangle. The coefficients are not magic; they are *the sizes of these bins*.

Pascal's recurrence: include the last factor, or don't

Now look at the addition rule that builds the triangle:

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

You can prove this algebraically by hammering on the factorial formula, but the combinatorial proof is much more satisfying — and it shows you why it must be true.

You want to count the ways to pick k items out of n. Single out one specific item — call it the last one. Every valid k-subset either contains the last item or does not. These two cases are mutually exclusive (no subset can both contain and not contain the same item), and together they cover every possibility (a subset must do exactly one of the two). So:

\binom{n}{k} = (\text{subsets that include the last item}) + (\text{subsets that exclude the last item})

Case A: include the last item. You have already committed to one of your k picks (the last item). You still need k-1 more, and you must pick them from the other n-1 items. The number of ways is \binom{n-1}{k-1}.

Case B: exclude the last item. You still need all k picks, and you must take them from the other n-1 items only (the last item is off-limits). The number of ways is \binom{n-1}{k}.

Add the two:

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

Why does this prove Pascal's recurrence — really? Because every k-subset of n items is counted exactly once by the right-hand side. If the subset contains the last item, it lives in case A and is counted by \binom{n-1}{k-1}. If not, it lives in case B and is counted by \binom{n-1}{k}. No subset is missed; no subset is counted twice. Sum equals total. The recurrence is forced by the disjoint-and-exhaustive logic — there is no other answer it could give.

In Pascal's triangle, "the entry in row n at position k" sits directly below "the entry in row n-1 at position k-1" (its upper-left neighbour) and "the entry in row n-1 at position k" (its upper-right neighbour). The "include the last factor / exclude the last factor" split is precisely the geometric fact that each entry has two parents above it. The addition rule is the recurrence; the recurrence is the include/exclude choice; the include/exclude choice is forced by basic logic. Algebra, geometry, and combinatorics are saying the same thing.

Concrete check: row 3 = 1, 3, 3, 1 by direct counting

Take n = 3 and just list. (a+b)^3 = (a+b)(a+b)(a+b). There are 2^3 = 8 raw picks. Write each one as a string of three letters ("which side of each bracket did you pick?"):

Pick Number of bs Monomial
aaa 0 a^3
aab 1 a^2 b
aba 1 a^2 b
baa 1 a^2 b
abb 2 a b^2
bab 2 a b^2
bba 2 a b^2
bbb 3 b^3

Sorting into bins by the number of bs: one a^3, three a^2 b, three a b^2, one b^3. So

(a+b)^3 = a^3 + 3 a^2 b + 3 a b^2 + b^3

The coefficients 1, 3, 3, 1 are literally the sizes of the bins. Row 3 of Pascal's triangle is the same list because Pascal's triangle is the lookup table for those bin sizes — and that, again, is what \binom{3}{k} means.

Notice that the bin sizes match the recurrence too. From row 2 — 1 \;\; 2 \;\; 1 — the entry "3" in row 3, position 1 comes from 1 + 2. In include/exclude language: the three picks with one b (aab, aba, baa) split into "the last letter is b" (just aab, one way — that is the \binom{2}{0} = 1 from row 2) and "the last letter is a" (aba and baa, two ways — that is the \binom{2}{1} = 2 from row 2). One plus two equals three. The recurrence is doing the bookkeeping.

Worked examples

Example 1 — Row 4 derived by combinatorial counting

You want the coefficients of (a+b)^4. There are four binomial factors, and the coefficient of a^{4-k} b^k counts the ways to pick k of the four factors to contribute a b.

  • k = 0: you pick zero of the four factors to give a b. There is exactly 1 way to do that — pick none. So \binom{4}{0} = 1.
  • k = 1: pick which one of four gives a b. There are 4 ways (factor 1, 2, 3, or 4). So \binom{4}{1} = 4.
  • k = 2: pick which two of four give bs. The pairs are \{1,2\}, \{1,3\}, \{1,4\}, \{2,3\}, \{2,4\}, \{3,4\} — 6 pairs. So \binom{4}{2} = 6.
  • k = 3: pick which three of four give bs. Equivalently, pick the one that gives an a — and there are 4 ways. So \binom{4}{3} = 4.
  • k = 4: pick all four to give bs. 1 way. So \binom{4}{4} = 1.

Why does \binom{4}{3} = 4 equal \binom{4}{1} = 4? Because choosing which 3 get b is the same as choosing which 1 gets a — every pick uniquely determines its complement. This is the symmetry \binom{n}{k} = \binom{n}{n-k}, and it is why Pascal's triangle is left–right symmetric.

So row 4 reads 1, 4, 6, 4, 1, and

(a+b)^4 = a^4 + 4 a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4

Spot-check at a = b = 1: 1 + 4 + 6 + 4 + 1 = 16 = 2^4, which is the count of all picks before binning. Confirmed.

Example 2 — Row 5 from row 4 by Pascal's recurrence

Row 4 is 1 \;\; 4 \;\; 6 \;\; 4 \;\; 1. Using \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}, row 5 is built by summing adjacent pairs from row 4 (with implicit 0s on either end so the leading and trailing 1s are preserved):

\underbrace{1}_{0+1} \quad \underbrace{1+4}_{5} \quad \underbrace{4+6}_{10} \quad \underbrace{6+4}_{10} \quad \underbrace{4+1}_{5} \quad \underbrace{1}_{1+0}

So row 5 is 1, 5, 10, 10, 5, 1.

Why does this addition give the right answer? Take the middle entry \binom{5}{2} = 10. It counts the ways to pick 2 of 5 factors. Split by the last factor: either it is one of the two b-picks (then choose 1 more from the other 4, giving \binom{4}{1} = 4 ways) or it is not (then choose both from the other 4, giving \binom{4}{2} = 6 ways). 4 + 6 = 10, exactly as the addition predicts.

So (a+b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5, with coefficients 1, 5, 10, 10, 5, 1. The triangle was built by addition, but every addition was an instance of the include/exclude split.

Example 3 — $\binom{5}{2}$ via the formula matches the row

The closed-form formula for the binomial coefficient is

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

For n = 5, k = 2:

\binom{5}{2} = \frac{5!}{2! \cdot 3!} = \frac{120}{2 \cdot 6} = \frac{120}{12} = 10

And the third entry of row 5 (positions 0, 1, \mathbf{2}, 3, 4, 5) is also 10 — matched.

Why does \frac{n!}{k!(n-k)!} count the same thing? You can list the k picks in n!/(n-k)! different orders if order matters (pick the first, then the second, ..., then the k-th, with the count shrinking each time). But the bins above don't care which order you picked the bs — only which k factors gave a b. Each subset of size k corresponds to k! ordered picks, so divide by k! to remove the ordering. That is the formula.

So the formula and the triangle agree, as they must — both compute the same count. The triangle is a fast lookup; the formula is the engine. For row 5 both are easy. For row 25 the formula is the only sane way; for row 7 in your head, the triangle wins. They are two presentations of one number.

What this gives you

Once you internalise that the coefficients are counts and the recurrence is a yes/no question about the last item, you no longer need to remember Pascal's triangle as a list of magic numbers. You can rebuild it from scratch in fifteen seconds, you can derive the symmetry \binom{n}{k} = \binom{n}{n-k} on the spot (by the complement argument from Example 1), and you have the same combinatorial reflex that makes harder identities like the hockey-stick (\sum_{i=k}^{n} \binom{i}{k} = \binom{n+1}{k+1}) and Vandermonde's (\sum_{j} \binom{m}{j}\binom{n}{k-j} = \binom{m+n}{k}) feel obvious instead of mysterious. They are all the same trick — count the same thing two ways, set the answers equal — applied to slightly different counting questions.

This is also the road into probability. The binomial distribution — the chance of k heads in n coin flips, or the chance of k sixes in n throws of a die, or the chance of k out of n Diwali candles staying lit through the night — is built directly on these same Pascal's-triangle counts. Once you see them as counts, the formulas you meet later in JEE and beyond stop being a list of things to memorise and start being a small handful of ideas applied over and over.

References

  1. Algebraic identities — the parent article. (a+b)^2 and (a+b)^3 are the smallest non-trivial cases of what this argument proves in general.
  2. Pascal's triangle: click a row, watch (a+b)ⁿ pop out — the companion visualisation, with a clickable interactive triangle.
  3. Wikipedia: Binomial coefficient — full discussion of \binom{n}{k}, including the combinatorial and algebraic definitions.
  4. Wikipedia: Pascal's rule — the recurrence \binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k} in its own right, with both proofs.
  5. NCERT Class 11 Mathematics, Chapter 7: Binomial Theorem — the standard Indian school treatment, which assumes the recurrence and uses it computationally.
  6. Cut-the-Knot: combinatorial proof of Pascal's identity — clean exposition of the include/exclude argument with several worked subset examples.