MISSION · Algebra
Pascal's Triangle and the Binomial Theorem
Where the coefficients come from, and why they're also combinations
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The idea in one sentence
The coefficients in \((a + b)^n\) are the \(n\)th row of Pascal’s triangle, and they are there because they count the ways of choosing terms.
Multiply it out once
\((a+b)^2 = a^2 + 2ab + b^2\) \((a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3\) \((a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4\)
Two patterns run through all of them. The powers of \(a\) count down while the powers of \(b\) count up, and every term’s exponents total \(n\). The coefficients are \(1, 2, 1\) then \(1, 3, 3, 1\) then \(1, 4, 6, 4, 1\).
Pascal’s triangle
Start with 1. Each entry below is the sum of the two above it:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
Row \(n\) gives the coefficients for \((a+b)^n\). Read off row 5 and you have \((a+b)^5\) without multiplying anything.
The addition rule isn’t a coincidence either. To get a term in the next expansion you can arrive from either of two terms in the current one, so the counts add.
The coefficients are combinations
\[\binom{n}{r} = \frac{n!}{r!\,(n-r)!}\]The coefficient of \(a^{n-r}b^r\) is exactly \(\binom{n}{r}\) — the number of ways of choosing \(r\) items from \(n\).
Here’s why. \((a+b)^4\) is four brackets multiplied together, and expanding means picking either \(a\) or \(b\) from each. To get \(a^2b^2\) you must pick \(b\) from exactly two of the four brackets, and there are \(\binom{4}{2} = 6\) ways to choose which two. That’s the 6 in row 4.
So the binomial theorem and the counting formula are the same fact seen from two sides, and Pascal’s triangle is a table of combinations.
\[(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\]Using it for one term
Most exam questions want a single term rather than the whole expansion, which is where the formula beats the triangle.
The \(x^3\) term of \((2x + 3)^5\): you need \(r = 2\), giving \(\binom{5}{2}(2x)^3(3)^2 = 10 \cdot 8x^3 \cdot 9 = 720x^3\).
Note that the coefficients inside the brackets get raised too. Forgetting to cube the 2 is the standard error.
Other things hiding in the triangle
The rows sum to powers of 2. The diagonals give the counting numbers, then the triangular numbers, then the tetrahedral ones. Colour the odd entries and the Sierpiński triangle appears. And the rows, scaled, approach the bell curve — which is why the binomial distribution tends towards the normal one.
Common mix-ups
- Using the wrong row. Row \(n\) starts with 1 and then \(n\); the top row is row 0.
- Ignoring coefficients inside the bracket. \((2x)^3 = 8x^3\).
- Losing alternating signs in \((a - b)^n\), where odd powers of \(b\) go negative.
- Mixing up \(r\) and \(n - r\), though the symmetry of the triangle often hides it.
Check yourself
Expand \((x + 2)^3\), and give the coefficient of \(x^2\) in \((x + 3)^4\).
Row 3 is \(1, 3, 3, 1\): \(x^3 + 6x^2 + 12x + 8\). For the second, \(\binom{4}{2}x^2 3^2 = 6 \cdot 9 = 54\).
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