Expanding $(a+b)^2$ or $(a+b)^3$ by hand is easy, but $(a+b)^{10}$ would mean multiplying ten brackets together. The Binomial Theorem gives the entire expansion of $(a+b)^n$ for any positive integer $n$ in one stroke, without doing the multiplication.
The numbers $\binom{n}{r}={}^{n}C_r=\dfrac{n!}{r!\,(n-r)!}$ are the binomial coefficients. Read the expansion column by column: the power of $a$ falls from $n$ down to $0$, the power of $b$ rises from $0$ up to $n$, and in every single term the two powers add to $n$. There are exactly $n+1$ terms.
Properties worth memorising: The expansion is symmetric because $\binom{n}{r}=\binom{n}{n-r}$, so coefficients read the same forwards and backwards. Putting $a=b=1$ gives $\binom{n}{0}+\binom{n}{1}+\cdots+\binom{n}{n}=2^n$ (the sum of all coefficients), while $a=1,\,b=-1$ gives the alternating sum $\binom{n}{0}-\binom{n}{1}+\cdots=0$. A handy special case is $(1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^{r}$.
Pascal's Triangle generates these coefficients without any factorials. Each row begins and ends in $1$, and every interior number is the sum of the two numbers diagonally above it — the identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$, known as Pascal's rule.
| $n$ | Coefficients (row of Pascal's triangle) | Row sum $=2^n$ | ||||||
|---|---|---|---|---|---|---|---|---|
| $0$ | $1$ | $1$ | ||||||
| $1$ | $1\quad 1$ | $2$ | ||||||
| $2$ | $1\quad 2\quad 1$ | $4$ | ||||||
| $3$ | $1\quad 3\quad 3\quad 1$ | $8$ | ||||||
| $4$ | $1\quad 4\quad 6\quad 4\quad 1$ | $16$ | ||||||
| $5$ | $1\quad 5\quad 10\quad 10\quad 5\quad 1$ | $32$ | ||||||
Deeper Insight — why the coefficients are exactly ${}^{n}C_r$: The binomial coefficient is not a mysterious lookup number; it is a count. When you expand $(a+b)^n=(a+b)(a+b)\cdots(a+b)$ with $n$ brackets, every term in the product is formed by choosing either $a$ or $b$ from each bracket. A term $a^{n-r}b^{r}$ arises whenever you pick $b$ from exactly $r$ of the $n$ brackets and $a$ from the rest — and the number of ways to choose those $r$ brackets is precisely $\binom{n}{r}$, the number of $r$-element subsets of an $n$-element set. This is why Pascal's identity $\binom{n}{r}=\binom{n-1}{r-1}+\binom{n-1}{r}$ holds: a subset either contains a fixed element (choose the other $r-1$ from $n-1$) or it does not (choose all $r$ from $n-1$). The same subset-counting also explains why all the coefficients add to $2^n$: that is the total number of subsets of an $n$-set. Seeing algebra and counting as two views of one idea makes the whole chapter feel inevitable rather than memorised.