In short

Row n of Pascal's triangle is the list of coefficients of (a+b)^n. The triangle starts with a 1 at the top, every other entry is the sum of the two numbers directly above it, and the k-th entry of row n is exactly \binom{n}{k} — the number of ways to pick k "b"s out of n binomial factors. Click any row in the widget below; the matching expansion of (a+b)^n appears with the row's numbers slotted in. One picture, two fields: this is the bridge from algebra to combinatorics.

You already know (a+b)^2 = a^2 + 2ab + b^2 and (a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3. The coefficients are 1, 2, 1 and 1, 3, 3, 1. What about (a+b)^7? You could multiply (a+b) by itself seven times — that takes a while and you will probably mess up a sign. Or you could write down a small triangle of numbers in about ten seconds and read the coefficients straight off row 7.

That triangle is Pascal's triangle. It is named after Blaise Pascal, but the Indian mathematician Pingala wrote about it more than two thousand years ago, and it appears in Halayudha's commentary, in the work of Varahamihira, and in 13th-century China as Yang Hui's triangle. It is the single most useful number pattern you will meet at school, and it does two jobs at once — it gives you the binomial expansion and it counts how many ways to choose k things out of n.

The triangle, and the rule that builds it

Start with row 0: just a single 1. Below it, row 1: 1 \;\; 1. Each new row starts and ends with a 1, and every entry in between is the sum of the two entries directly above it. So row 2 is 1 \;\; (1+1) \;\; 1 = 1 \;\; 2 \;\; 1. Row 3 is 1 \;\; (1+2) \;\; (2+1) \;\; 1 = 1 \;\; 3 \;\; 3 \;\; 1. Keep going forever.

Why does the "sum of the two above" rule work? When you go from (a+b)^n to (a+b)^{n+1}, you multiply by one more (a+b). A term a^{n-k}b^k in row n contributes once to a^{n-k+1}b^k (when you pick the a from the new factor) and once to a^{n-k}b^{k+1} (when you pick the b). So each entry in the next row gets contributions from its two upstairs neighbours — that is exactly Pascal's recurrence \binom{n+1}{k} = \binom{n}{k-1} + \binom{n}{k}.

Click a row

The triangle below shows rows 0 through 7. Click any row and the corresponding expansion of (a+b)^n will appear underneath, with the row's numbers slotted in as coefficients.

Click any row above to see the expansion of (a + b)ⁿ.
Pascal's triangle, rows $0$ through $7$. Each entry is the sum of the two directly above it. Click a row to read off $(a+b)^n$ with that row's coefficients filled in.

Notice what is happening as you click. Row 0 gives (a+b)^0 = 1 — the empty product, just a 1. Row 1 gives (a+b)^1 = a + b. Row 2 gives the familiar a^2 + 2ab + b^2. Row 7 — which would be a nightmare to expand by hand — pops out as a^7 + 7a^6b + 21a^5b^2 + 35a^4b^3 + 35a^3b^4 + 21a^2b^5 + 7ab^6 + b^7 in one click. The exponent of a starts at n on the left and decreases; the exponent of b starts at 0 and increases; the two exponents always sum to n. That is the binomial theorem:

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

The \binom{n}{k} symbol — read "n choose k" — is just the k-th entry of row n.

Why does the row count binomial coefficients?

Here is the bit that turns Pascal's triangle from a numerical curiosity into the bridge between algebra and combinatorics. When you write (a+b)^n, you are multiplying n identical copies of (a+b):

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

To expand, you walk through the n brackets and from each one you pick either an a or a b. Multiply the picks together, and you get one term. There are 2^n ways to make those picks, so there are 2^n terms before you collect like ones. Now group the picks by how many bs you chose. If you chose k of the bs (and therefore n-k of the as), the resulting term is a^{n-k}b^k — the same monomial regardless of which k brackets gave you the b.

Why does the coefficient equal \binom{n}{k}? It counts the number of ways to choose which k of the n brackets contribute a b. That is exactly the definition of "n choose k": pick k items from n without caring about order. So the coefficient of a^{n-k}b^k is \binom{n}{k}, and that number lives in row n, position k of Pascal's triangle.

This is why the same numbers appear in two completely different-looking problems. "How many ways can I pick 2 chocolates from 4 different ones in the box?" and "What is the coefficient of a^2 b^2 in (a+b)^4?" have the same answer — both are \binom{4}{2} = 6 — because both ask: in how many ways can you tag 2 slots out of 4? Pascal's triangle is the lookup table for both questions.

Worked examples

Example 1 — Reading row 4 off the triangle, with the addition shown

Row 4 is 1 \;\; 4 \;\; 6 \;\; 4 \;\; 1. Where did each number come from? From row 3, which is 1 \;\; 3 \;\; 3 \;\; 1, by adding adjacent pairs:

\underbrace{1}_{1} \quad \underbrace{1+3}_{4} \quad \underbrace{3+3}_{6} \quad \underbrace{3+1}_{4} \quad \underbrace{1}_{1}

The middle 6 is the sum of the two 3s above it. The 4s come from 1+3 and 3+1. Why is the second entry 4 = 1+3? Because to get a^3 b in (a+b)^4 = (a+b)^3 \cdot (a+b), you can either take the a^3 term from (a+b)^3 (coefficient 1) and pair it with the b from the new factor, or take the a^2 b term (coefficient 3) and pair it with the a. The two ways add: 1+3 = 4.

So row 4 gives

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

Spot-check with a = b = 1: 1 + 4 + 6 + 4 + 1 = 16 = 2^4. The sum of any row n is always 2^n, because that is what (1+1)^n evaluates to.

Example 2 — Two chocolates from a box of four (the same number, twice)

Suppose your friend brings a box of four different Diwali chocolates — kaju katli, soan papdi, gulab jamun, and rasgulla — and tells you to pick any two. In how many different pairs can you walk away?

You are choosing 2 items out of 4. The answer is \binom{4}{2}, which lives at position k=2 of row n=4 in Pascal's triangle: 1 \;\; 4 \;\; \mathbf{6} \;\; 4 \;\; 1. So six different pairs:

\{\text{kaju, soan}\}, \{\text{kaju, gulab}\}, \{\text{kaju, rasgulla}\}, \{\text{soan, gulab}\}, \{\text{soan, rasgulla}\}, \{\text{gulab, rasgulla}\}

Now look at (a+b)^4 = a^4 + 4a^3 b + 6 a^2 b^2 + 4 a b^3 + b^4. The coefficient of a^2 b^2 is also 6. Why the same number? Because writing (a+b)^4 = (a+b)(a+b)(a+b)(a+b) and asking "in how many ways can I get exactly two bs from the four brackets?" is the same question as "in how many ways can I tag 2 of 4 chocolates?" — both are \binom{4}{2}. Algebra and combinatorics are the same problem in disguise.

This identical-number trick is the entire reason probability and combinatorics use binomial coefficients so heavily. The number of ways a coin lands heads k times in n flips is \binom{n}{k} — the same row-n, position-k entry of Pascal's triangle.

Example 3 — Expanding $(2x + 3)^5$ using row 5

Row 5 of Pascal's triangle is 1 \;\; 5 \;\; 10 \;\; 10 \;\; 5 \;\; 1. To expand (2x+3)^5, treat a = 2x and b = 3, and slot them into the binomial template:

(2x+3)^5 = \binom{5}{0}(2x)^5 + \binom{5}{1}(2x)^4(3) + \binom{5}{2}(2x)^3(3)^2 + \binom{5}{3}(2x)^2(3)^3 + \binom{5}{4}(2x)(3)^4 + \binom{5}{5}(3)^5

Now compute each term separately. Why split it like this? Each piece has three independent things to handle: the row-5 Pascal coefficient, the power of 2x, and the power of 3. Doing them one at a time keeps the arithmetic clean.

  • 1 \cdot (2x)^5 = 32 x^5
  • 5 \cdot (2x)^4 \cdot 3 = 5 \cdot 16 x^4 \cdot 3 = 240 x^4
  • 10 \cdot (2x)^3 \cdot 9 = 10 \cdot 8 x^3 \cdot 9 = 720 x^3
  • 10 \cdot (2x)^2 \cdot 27 = 10 \cdot 4 x^2 \cdot 27 = 1080 x^2
  • 5 \cdot (2x) \cdot 81 = 810 x
  • 1 \cdot 243 = 243

Adding the lot:

(2x+3)^5 = 32 x^5 + 240 x^4 + 720 x^3 + 1080 x^2 + 810 x + 243

Spot-check at x = 1: (2+3)^5 = 5^5 = 3125. The sum on the right is 32 + 240 + 720 + 1080 + 810 + 243 = 3125. Confirmed.

Without Pascal's triangle, you would multiply (2x+3) by itself five times — six full distributions, dozens of like-terms to collect. With the triangle, six independent multiplications and a final add. This is exactly the kind of speed JEE problems demand.

The bridge to combinatorics, probability, and JEE

Pascal's triangle is the single object where algebra and combinatorics shake hands. Algebra says: row n gives the coefficients in (a+b)^n. Combinatorics says: the same row counts the ways to pick subsets of various sizes from a set of n items. Probability says: if you flip a fair coin n times, the chance of exactly k heads is \binom{n}{k} / 2^n — Pascal-row-n entry divided by the row's total.

In JEE, the binomial theorem itself is one of the highest-yield chapters. Many problems boil down to "find the coefficient of x^7 in (1 + x)^{10}(1 - x)^{15}", or "find the term independent of x in (x^2 + 1/x)^{12}" — and you solve them by reaching into Pascal's triangle for the right entry. In probability and statistics, the binomial distribution has Pascal's row n literally as its shape, and the normal curve that physicists use for measurement errors is the limiting smooth shape Pascal's rows take as n grows large. So the same picture you can sketch on a piece of notebook paper is the reason your weather app can quote rainfall probabilities.

Once you can read (a+b)^n off Pascal's triangle in your head, you are also reading off \binom{n}{k}, also reading off coin-flip probabilities, also reading off the Indian-mathematician-Pingala-was-here pattern from over two thousand years ago. One number triangle, three different fields, no memorisation — just the rule "each entry is the sum of the two above it," repeated.

References

  1. Algebraic identities — the parent article, and where (a+b)^2, (a+b)^3 live as the first two non-trivial cases.
  2. Geometric proof of (a+b)³: eight-piece cube dissection — the row-3 coefficients 1, 3, 3, 1 as physical pieces of a cube.
  3. Wikipedia: Pascal's triangle — full history, including Pingala, Halayudha, and Yang Hui.
  4. Wikipedia: Binomial theorem — the general statement (a+b)^n = \sum \binom{n}{k} a^{n-k} b^k.
  5. NCERT Class 11 Mathematics, Chapter 7: Binomial Theorem — the standard Indian school treatment, with JEE-style worked problems.
  6. Brilliant: Pascal's triangle and the binomial theorem — interactive practice with combinatorial identities.