The Binomial Theorem Advanced. This lesson goes beyond core Algebra II. You can skip it.

Learning goals

State the Binomial Theorem, where $\binom{n}{k}$ counts the ways factors donate $y$
Expand $(A+B)^n$ using row $n$ of Pascal's triangle, and alternate signs for a difference
Build Pascal's triangle from the addition rule $\binom{n}{k}=\binom{n-1}{k-1}+\binom{n-1}{k}$
Pick out one term of an expansion by solving for $k$ from its exponent

Multiplying out is really choosing

A power like $(x + y)^7$ is not a new kind of object. It is shorthand for a product of seven identical factors:

(x + y)^7 = \underbrace{(x + y)(x + y)\cdots(x + y)}_{7 \text{ factors}}.

So expanding it is just multiplying out a product of sums, and the distributive law has told you how to do that since the beginning. Take one term from each factor, multiply your picks together, and add up the result over every possible way of picking. Nothing else appears in the product, and nothing is left out.

Here every factor is the same, and each one offers exactly two things: an $x$ or a $y$ . Watch what that means for the two smallest cases before asking about $(x+y)^7$ or anything bigger.

The four picks for $n = 2$ are $xx$ , $xy$ , $yx$ , and $yy$ , so

(x + y)^2 = x^2 + xy + yx + y^2 = x^2 + 2xy + y^2.

That $2$ is not decoration. It counts the two picks that took a single $y$ , one from the first factor and one from the second. Now do $n = 3$ , where there are $2^3 = 8$ picks:

Every product in the expansion of a cube is decided by which of the three factors donate a y. Three of the eight products come out as x squared times y, and that is exactly why its coefficient is 3.

Collecting the eight products by pile gives

(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3,

and the $3$ in front of $x^2y$ is not a mystery number. It is the count of ways to pick which one of the three factors donated its $y$ , and there are three factors to pick from.

The doubling never stops. Build the list of picks one factor at a time. Each way of picking from the first $n - 1$ factors turns into exactly two ways once an $n$ -th factor arrives, one taking its $x$ and one taking its $y$ . No other way is possible, so every new factor doubles the list: $4$ picks for $n = 2$ , $8$ for $n = 3$ , and in general $2^n$ picks for $(x + y)^n$ . Beginning from the single empty pick, $n$ doublings give $2^n$ ways in all. The expansion of $(x + y)^n$ is therefore a sum of $2^n$ products before a single like term is collected.

What does one of those products look like in general? Suppose your picks took the $y$ from exactly $k$ of the factors. Then they took the $x$ from the other $n - k$ , and since multiplication does not care about order, the product is $x^{n-k} y^{k}$ . That is the whole shape of the answer. Every one of the $2^n$ products is $x^{n-k}y^k$ for some whole number $k$ between $0$ and $n$ . The exponents always add up to $n$ , and no other kind of term can possibly appear. What is left to find is how many products land on each pile.

That observation deserves a name, because it is the entire lesson.

Definition. For whole numbers $k$ and $n$ with $0 \le k \le n$ , write $\binom{n}{k}$ , read ” $n$ choose $k$ ”, for the number of ways to choose which $k$ of the $n$ factors donate their $y$ .

Notice what this definition is and is not. It is not a formula you have to trust. It is a description of something you can count, and you have already counted four of them by hand:

\binom{3}{0} = 1, \qquad \binom{3}{1} = 3, \qquad \binom{3}{2} = 3, \qquad \binom{3}{3} = 1.

A single pick is completely described by the set of factors that gave a $y$ . So counting the picks with exactly $k$ of the $y$ ‘s and counting the sets of $k$ factors are the same job. That is why one number answers both questions.

The Binomial Theorem

The expansion has one term for each possible number of $y$ -picks, from $k = 0$ to $k = n$ :

(x + y)^n = \binom{n}{0}x^n + \binom{n}{1}x^{n-1}y + \binom{n}{2}x^{n-2}y^2 + \cdots + \binom{n}{n}y^n,

which is $n + 1$ terms, one for each value of $k$ from $0$ to $n$ .

That same list can be written compactly using sigma notation:

(x + y)^n \;=\; \sum_{k=0}^{n} \binom{n}{k}\, x^{n-k} y^{k}.

The $\sum_{k=0}^{n}$ means “add one term for every whole-number value of $k$ from $0$ through $n$ .” It is nothing more than shorthand for the sum you already wrote out above.

You have already done the work that proves this. Every product in the expansion has the form $x^{n-k}y^k$ for some $k$ from $0$ to $n$ , and $\binom{n}{k}$ counts exactly how many of the $2^n$ products land in that pile. Summing the piles for every $k$ therefore accounts for the whole expansion, which is the Binomial Theorem.

Before building a shortcut for the coefficients, use the theorem to fill in only the exponents of $(a + b)^4$ . There are $n + 1 = 5$ terms, one for each $k$ from $0$ to $4$ :

a^4, \quad \square\, a^3b, \quad \square\, a^2b^2, \quad \square\, ab^3, \quad b^4.

Notice the pattern even before a single coefficient is known: the exponent of $a$ falls by $1$ each term while the exponent of $b$ climbs by $1$ , and every pair still adds to $4$ . The two ends are already filled in, since $\binom{4}{0} = \binom{4}{4} = 1$ . The three missing boxes are $\binom{4}{1}$ , $\binom{4}{2}$ , and $\binom{4}{3}$ , exactly the numbers Pascal’s triangle is about to hand you without any more counting by hand.

Pascal’s triangle, built from addition alone

The theorem is only as useful as your ability to compute $\binom{n}{k}$ , and so far the only method on offer is to list every selection by hand. For $(x + y)^{10}$ that means writing out $2^{10} = 1024$ strings, which is not a method, it is a punishment. One short argument replaces the listing with addition.

Start with the two easy counts. There is exactly one way for no factor to donate a $y$ (every factor donates its $x$ ), and exactly one way for all of them to donate a $y$ . So for every $n$ ,

\binom{n}{0} = 1 \qquad\text{and}\qquad \binom{n}{n} = 1.

One more fact comes free before you even build the rest of the triangle. Choosing which $k$ factors donate a $y$ is the same act as choosing which $n - k$ factors donate an $x$ : fix one selection and the other is decided. Two descriptions of the same act must have the same count, so

\binom{n}{k} = \binom{n}{n-k}.

Every row of the triangle you are about to build will read the same forwards and backwards because of this.

Everything in between the edges follows from a single rule.

Pascal's rule: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ #

Start with a concrete case: choosing $2$ of $5$ labeled factors to donate a $y$ . Look at factor $5$ alone. A selection either includes factor $5$ or it does not, and never both.

If it includes factor $5$ , factor $5$ already supplies one of the two $y$ ‘s, and the other comes from the remaining $4$ factors: $\binom{4}{1} = 4$ ways. If it does not include factor $5$ , both $y$ ‘s must come from the remaining $4$ factors: $\binom{4}{2} = 6$ ways. Listing them out confirms the split:

\underbrace{\{1,5\},\ \{2,5\},\ \{3,5\},\ \{4,5\}}_{\text{4 pairs that include factor 5}}

\underbrace{\{1,2\},\ \{1,3\},\ \{1,4\},\ \{2,3\},\ \{2,4\},\ \{3,4\}}_{\text{6 pairs that do not}}

Every selection falls into exactly one of these two groups, so

\binom{5}{2} = \binom{4}{1} + \binom{4}{2} = 4 + 6 = 10.

Nothing about that split depended on the specific numbers $5$ and $2$ . Take any $n \ge 2$ and any interior index $k$ with $1 \le k \le n - 1$ , and look at the last of the $n$ factors, factor $n$ . A selection of $k$ factors to donate a $y$ either uses factor $n$ or it does not.

If it uses factor $n$ , the other $k - 1$ $y$ ‘s come from the first $n - 1$ factors, giving $\binom{n-1}{k-1}$ ways. If it does not, all $k$ $y$ ‘s come from the first $n - 1$ factors, giving $\binom{n-1}{k}$ ways. The two groups are disjoint and together they are every selection counted by $\binom{n}{k}$ , so the total is the sum of the two counts:

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

Both symbols on the right stay legal: $1 \le k$ makes $k - 1 \ge 0$ , and $k \le n - 1$ keeps $\binom{n-1}{k}$ inside its own range $0 \le k \le n - 1$ .

Stack the counts $\binom{n}{0}, \binom{n}{1}, \ldots, \binom{n}{n}$ as row $n$ of a triangle, with the $1$ ‘s from the edge values anchoring both ends. Pascal’s rule says each interior entry is the sum of the two entries directly above it.

Pascal's triangle, rows 0 through 6. The edges are 1 because there is exactly one way to take no y at all and one way to take a y from every factor. Every interior entry is the sum of the two above it, so 4 + 6 = 10.

Row $n$ of the triangle is the complete list of coefficients for $(x + y)^n$ , and row $3$ is $1,\ 3,\ 3,\ 1$ , exactly what the hand count produced. Two more rows reach everything you will need below:

\begin{aligned} n = 7:&\quad 1,\ 7,\ 21,\ 35,\ 35,\ 21,\ 7,\ 1\\ n = 8:&\quad 1,\ 8,\ 28,\ 56,\ 70,\ 56,\ 28,\ 8,\ 1 \end{aligned}

Before reading any more rows off the page, build one entry yourself with the addition rule.

Check your understanding

Row $6$ of Pascal's triangle is $1,\ 6,\ 15,\ 20,\ 15,\ 6,\ 1$ . Using Pascal's rule, what is the entry at $k = 2$ in row $7$ ?

15

20

21

35

Two habits keep you on the right row. Row $n$ always has $n + 1$ entries, and its second entry is always $n$ , since $\binom{n}{1} = n$ counts the $n$ single factors that could have donated the lone $y$ .

Check your understanding

Use Pascal's triangle. What is the coefficient of $x^2y^3$ in the expansion of $(x + y)^5$ ?

5

10

20

15

Expanding a power

To expand $(A + B)^n$ , write down row $n$ of the triangle and march $k$ from $0$ to $n$ , giving $A$ the exponent $n - k$ and $B$ the exponent $k$ . The one thing to guard is that $A$ and $B$ are whole terms, coefficients included: if $A = 2a$ , then $A^3$ is $(2a)^3 = 8a^3$ , not $2a^3$ .

Once you have expanded, there is a quick way to catch an arithmetic slip. Setting every variable to $1$ turns the expansion into the sum of its coefficients, while the unexpanded side stays easy to evaluate by hand. If the two sides disagree, an error is guaranteed. If they agree, that is reassuring, but it does not certify every coefficient: two mistakes can still cancel each other out and leave the sums equal.

Worked example 1 Expand $(2a + 3)^4$

Here $A = 2a$ , $B = 3$ , and $n = 4$ , so the coefficients come from row $4$ , which is $1,\ 4,\ 6,\ 4,\ 1$ . Write the five terms with $(2a)$ losing a power and $3$ gaining one:

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

Now evaluate each piece, keeping the whole of $2a$ inside its power:

(2a)^4 = 16a^4, \qquad 4(8a^3)(3) = 96a^3, \qquad 6(4a^2)(9) = 216a^2,

4(2a)(27) = 216a, \qquad 3^4 = 81.

So the expansion is

(2a + 3)^4 = 16a^4 + 96a^3 + 216a^2 + 216a + 81.

Check it by setting $a = 1$ . The left side is $(2 + 3)^4 = 625$ , and the right side is $16 + 96 + 216 + 216 + 81 = 625$ . The check passes: if the two sides had disagreed, that would prove an error, but agreement here is reassuring rather than a guarantee that every coefficient is correct.

Powers of a difference

Nothing new is needed for $(x - y)^n$ , because a difference is a sum in disguise. Write $x - y$ as $x + (-y)$ and let the theorem do the work:

(x - y)^n = \big(x + (-y)\big)^n = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} (-y)^k.

The only thing to simplify is $(-y)^k = (-1)^k y^k$ , which drags one minus sign along for each factor that donated a $y$ . So

(x - y)^n = \sum_{k=0}^{n} (-1)^k \binom{n}{k}\, x^{n-k} y^{k},

and since $\binom{n}{k}$ is a count it is never zero or negative. The sign of the coefficient of $x^{n-k}y^k$ is therefore the sign of $(-1)^k$ alone: the coefficient is negative exactly when $k$ is odd, and positive exactly when $k$ is even. The signs alternate $+, -, +, -$ , starting from $+$ on the pure $x^n$ term.

Worked example 2 Expand $(x - 2)^4$

Take $A = x$ and $B = -2$ , so that the theorem applies to the sum $x + (-2)$ . Row $4$ is $1,\ 4,\ 6,\ 4,\ 1$ , and the powers of $-2$ climb from $k = 0$ to $k = 4$ :

(x - 2)^4 = x^4 + 4x^3(-2) + 6x^2(-2)^2 + 4x(-2)^3 + (-2)^4.

Evaluate the powers of $-2$ , which alternate in sign: $(-2)^1 = -2$ , $(-2)^2 = 4$ , $(-2)^3 = -8$ , and $(-2)^4 = 16$ . That gives

(x - 2)^4 = x^4 - 8x^3 + 24x^2 - 32x + 16.

The signs alternate because the odd powers of $-2$ are negative and the even ones are positive, which is the $(-1)^k$ from the derivation showing up in the arithmetic.

Check with $x = 1$ . The left side is $(1 - 2)^4 = 1$ , and the right side is $1 - 8 + 24 - 32 + 16 = 1$ . The check passes, which is reassuring, though it does not by itself guarantee every sign is correct: verify the alternating pattern directly from the powers of $-2$ , negative at the odd exponents and positive at the even ones.

Check your understanding

What is the coefficient of $x^4$ in the expansion of $(x - 1)^6$ ?

-20

20

15

-15

A closed form for the coefficients

Pascal’s triangle gives every coefficient you could want, but it makes you build all the rows below the one you need. If a problem asks for a coefficient in $(1 + x)^{20}$ , you do not want to write twenty rows. There is a closed-form formula that always produces the same number as counting the selections by hand.

First, notation. For a positive whole number $m$ , the factorial $m!$ is the product of every whole number from $m$ down to $1$ :

m! = m \times (m-1) \times \cdots \times 2 \times 1, \qquad\text{so}\qquad 4! = 4 \cdot 3 \cdot 2 \cdot 1 = 24.

The empty case is defined to be $0! = 1$ . That is a convention, chosen so the formula below gives $\binom{n}{0} = \binom{n}{n} = 1$ , matching the counts you already found by hand.

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

In practice you never expand the factorials in full. Cancel $(n-k)!$ against the tail of $n!$ first, which leaves $k$ factors on top and $k$ factors on the bottom.

Worked example 3 Find the coefficient of $x^3$ in $(1 + x)^{20}$

Take $A = 1$ and $B = x$ . Since $1^{20-k} = 1$ for every $k$ , the term holding $x^3$ is simply $\binom{20}{3}x^3$ , and the coefficient is $\binom{20}{3}$ .

Twenty rows of Pascal’s triangle would settle it, and so would the formula, in one line. Write out the factorials and cancel $17!$ against the tail of $20!$ :

\binom{20}{3} = \frac{20!}{3!\,17!} = \frac{20 \times 19 \times 18 \times 17!}{(3 \times 2 \times 1) \times 17!} = \frac{20 \times 19 \times 18}{6}.

Cancel before multiplying: $18 \div 6 = 3$ , so the product is $20 \times 19 \times 3$ .

\binom{20}{3} = 1140.

The coefficient of $x^3$ is $1140$ . Notice the shape of the shortcut: three factors on top counting down from $20$ , and $3!$ underneath. That pattern is what the cancellation always produces.

The same formula also explains the symmetry you spotted earlier, since swapping $k$ and $n - k$ swaps the two factorials in the denominator and leaves the value alone.

Picking out a single term

Expanding a whole power is rarely the point. The question that actually gets asked is “what is the coefficient of $x^{12}$ ”, and expanding twenty terms to read off one of them is wasted work. The theorem lets you go straight to it, because it hands you the general term:

T_k = \binom{n}{k}\, A^{\,n-k} B^{\,k}, \qquad k = 0, 1, 2, \ldots, n.

The method is always the same three steps. Write the general term for your $A$ and $B$ . Collect all the powers of the variable into a single exponent. Set that exponent equal to the one you want, solve for $k$ , and only then evaluate.

Worked example 4 Find the coefficient of $a^3b^5$ in $(2a - b)^8$

Here $A = 2a$ and $B = -b$ , with $n = 8$ . The general term is

T_k = \binom{8}{k} (2a)^{8-k} (-b)^{k} = \binom{8}{k}\, 2^{8-k} (-1)^k\, a^{8-k} b^{k}.

You want $b^5$ , so $k = 5$ . That also fixes the power of $a$ at $8 - 5 = 3$ , which matches the $a^3$ asked for, so such a term does exist. Now evaluate the three pieces. Row $8$ gives $\binom{8}{5} = 56$ , the power of $2$ is $2^{8-5} = 2^3 = 8$ , and $(-1)^5 = -1$ because $5$ is odd:

T_5 = 56 \cdot 8 \cdot (-1)\, a^3 b^5 = -448\,a^3b^5.

The coefficient is $-448$ . Two traps sat in that problem: the $2$ inside $2a$ had to be cubed along with the $a$ , and the odd $k$ made the sign negative.

Worked example 5 Find the coefficient of $x^6$ in $(2 + x^2)^5$

Here $A = 2$ , $B = x^2$ , and $n = 5$ . The general term is

T_k = \binom{5}{k}\, 2^{5-k}\, (x^2)^k = \binom{5}{k}\, 2^{5-k}\, x^{2k}.

You want $x^6$ , so set the exponent equal to $6$ and solve for $k$ :

2k = 6 \quad\Longrightarrow\quad k = 3.

Now evaluate. Row $5$ gives $\binom{5}{3} = 10$ , and the power of $2$ is $2^{5-3} = 2^2 = 4$ :

T_3 = 10 \cdot 4\, x^6 = 40x^6.

The coefficient is $40$ . Solving $2k = 6$ for $k$ is the whole method here: no negative exponents and no fractions, just one equation in $k$ .

Check your understanding

What is the coefficient of $x^3$ in the expansion of $(1 + 2x)^5$ ?

10

20

40

80

Common mistakes

Raising the variable but not its coefficient. In $(2a + 3)^4$ the term with $k = 1$ is $4(2a)^3(3) = 96a^3$ , not $4a^3(3)$ . Whatever sits inside the parentheses is the whole of $A$ , and the exponent lands on all of it.
Losing the alternating sign. In $(x - y)^n$ the coefficient of $x^{n-k}y^k$ carries a factor of $(-1)^k$ , so it is negative exactly when $k$ is odd. Rewrite the difference as $x + (-y)$ and let the minus sign travel inside the power, rather than trying to remember which terms it lands on.
Inventing terms whose exponents do not add to $n$ . Every term of $(x + y)^7$ has its two exponents summing to $7$ , because each of the seven factors donates exactly one letter. There is no $x^4y^2$ term to find, so a search for one has gone wrong before it started.

Key takeaways

$(x + y)^n$ means choosing an $x$ or a $y$ from each of $n$ identical factors, so every term is $x^{n-k}y^k$ . The coefficient $\binom{n}{k}$ counts the ways to choose which $k$ factors donate their $y$ .
The Binomial Theorem: $(x + y)^n = \displaystyle\sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ , an expansion with $n + 1$ terms. Also $\binom{n}{k} = \binom{n}{n-k}$ , and $\binom{n}{k} = \dfrac{n!}{k!\,(n-k)!}$ with $0! = 1$ .
Pascal’s rule: $\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-1}{k}$ builds the whole triangle from the two edges of $1$ ‘s. For $(x - y)^n$ the signs alternate, negative exactly when $k$ is odd.
To pick out one term: write the general term $\binom{n}{k}A^{n-k}B^k$ , collect the powers of the variable into one exponent, set it equal to the target exponent, and solve for $k$ .

Practice

Multiple Choice Questions (MCQ)

Progressively harder sets of questions. Each opens on its own page.

Level 1 Foundational 12 questions Start → Level 2 Intermediate 12 questions Start → Level 3 Challenge 12 questions Start →

Core practice

Practice problems at the level of the course, to be worked out on paper. Hints one at a time, then the answer or the full worked solution, with your progress kept in this browser.

More practice (optional)

Extra sets, as hard as the Challenge set. Each one opens on its own page.

More resources (optional)

Other explanations of this lesson, if you want a second take.

Binomial theorem (Math is Fun)
Pascal's triangle (Math is Fun)
Binomial theorem (OpenStax)
Binomial theorem (Art of Problem Solving)

Go deeper (optional)

You can skip this and keep going. Read it if you want to know more.

Why the factorial formula always agrees with the counting definition

The counting number $\binom{n}{k}$ equals $\dfrac{n!}{k!\,(n-k)!}$ #

Write $f(n, k) = \dfrac{n!}{k!\,(n-k)!}$ for $0 \le k \le n$ . We never assume $f$ counts anything. We claim $f$ always equals $\binom{n}{k}$ , because $f$ obeys the same two rules that build the triangle, and those rules leave no freedom at all. Run the whole argument once on $n = 6$ before writing it with letters.

The edges match, checked at $n = 6$ . Using $0! = 1$ ,

f(6, 0) = \frac{6!}{0!\,6!} = 1 \qquad\text{and}\qquad f(6, 6) = \frac{6!}{6!\,0!} = 1,

which are exactly the values of $\binom{6}{0}$ and $\binom{6}{6}$ .

Pascal’s rule holds for $f$ , checked at $n = 6$ , $k = 2$ . Add $f(5, 1)$ and $f(5, 2)$ and put them over the common denominator $2!\,4! = 48$ , multiplying the first fraction above and below by $k = 2$ and the second by $n - k = 4$ (deliberately different multipliers this time, so the two steps stay visibly distinct):

\begin{aligned} f(5,1) + f(5,2) &= \frac{5!}{1!\,4!} + \frac{5!}{2!\,3!} = \frac{2 \cdot 5!}{2!\,4!} + \frac{4 \cdot 5!}{2!\,4!} \\ &= \frac{240 + 480}{48} = 15. \end{aligned}

And directly, $f(6, 2) = \dfrac{6!}{2!\,4!} = \dfrac{720}{48} = 15$ . The two match.

The two rules determine every entry, checked through row $6$ . Row $0$ has $\binom{0}{0} = f(0,0) = 1$ . Using only the edge values and the addition rule just checked, build each row from the one above it, adding actual numbers rather than taking it on faith:

Row $1$ : edges only, $1, 1$ .
Row $2$ : edges $1, 1$ ; interior $\binom{2}{1} = \binom{1}{0} + \binom{1}{1} = 1 + 1 = 2$ . Row: $1, 2, 1$ .
Row $3$ : edges $1, 1$ ; interior $\binom{3}{1} = 1 + 2 = 3$ and $\binom{3}{2} = 2 + 1 = 3$ . Row: $1, 3, 3, 1$ .
Row $4$ : edges $1, 1$ ; interior $\binom{4}{1} = 1 + 3 = 4$ , $\binom{4}{2} = 3 + 3 = 6$ , $\binom{4}{3} = 3 + 1 = 4$ . Row: $1, 4, 6, 4, 1$ .
Row $5$ : edges $1, 1$ ; interior $\binom{5}{1} = 1 + 4 = 5$ , $\binom{5}{2} = 4 + 6 = 10$ , $\binom{5}{3} = 6 + 4 = 10$ , $\binom{5}{4} = 4 + 1 = 5$ . Row: $1, 5, 10, 10, 5, 1$ .
Row $6$ : edges $1, 1$ ; interior $\binom{6}{1} = 1 + 5 = 6$ , $\binom{6}{2} = 5 + 10 = 15$ , $\binom{6}{3} = 10 + 10 = 20$ , $\binom{6}{4} = 10 + 5 = 15$ , $\binom{6}{5} = 5 + 1 = 6$ . Row: $1, 6, 15, 20, 15, 6, 1$ .

Every one of those entries equals the matching value of $f$ ; in particular $\binom{6}{2} = 15$ matches the $f(6, 2) = 15$ found above. Nothing about this forcing used a special property of $6$ : the same edges-plus-addition argument would force agreement at row $7$ , row $8$ , and every row after that, one row at a time.

Now repeat the same argument in general, with $n$ and $k$ standing for an arbitrary row and index. Nothing below depends on $6$ having been the row checked above; the same three steps work starting from any row.

The edges match. Using $0! = 1$ ,

f(n, 0) = \frac{n!}{0!\,n!} = 1 \qquad\text{and}\qquad f(n, n) = \frac{n!}{n!\,0!} = 1,

which are exactly the values of $\binom{n}{0}$ and $\binom{n}{n}$ .

Pascal’s rule holds for $f$ . Take an interior index, $1 \le k \le n-1$ , so that every factorial below is defined. Add the two entries from the row above and put them over the common denominator $k!\,(n-k)!$ , multiplying the first fraction above and below by $k$ and the second by $n - k$ , exactly as in the numerical case:

\begin{aligned} f(n-1, k-1) + f(n-1, k) &= \frac{(n-1)!}{(k-1)!\,(n-k)!} + \frac{(n-1)!}{k!\,(n-1-k)!} \\ &= \frac{k\,(n-1)!}{k!\,(n-k)!} + \frac{(n-k)\,(n-1)!}{k!\,(n-k)!}. \end{aligned}

The two numerators now combine, and $k + (n - k) = n$ collapses them:

\frac{(n-1)!\,\big[k + (n-k)\big]}{k!\,(n-k)!} = \frac{n \cdot (n-1)!}{k!\,(n-k)!} = \frac{n!}{k!\,(n-k)!} = f(n, k).

The two rules determine everything. Row $0$ has the single entry $\binom{0}{0} = 1$ , and $f(0, 0) = 1$ as well, so the two agree there. Now suppose they agree on row $n - 1$ . In row $n$ the two end entries are $1$ for both, and every interior entry of both is the sum of the same two entries from row $n - 1$ . So the two agree on row $n$ as well. Working down one row at a time, exactly as the numerical case did through row $6$ , they agree on every row, and therefore

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

for all $0 \le k \le n$ . The counting definition and the formula are the same number.

The constant term of a binomial with negative exponents

The same three-step method works even when $B$ is not a positive power of the variable. Writing a $\dfrac{1}{x}$ term as $x^{-1}$ lets you add exponents instead of tracking fractions.

Worked example Find the constant term of $\left(x^2 + \dfrac{3}{x}\right)^{6}$

A constant term is the term whose power of $x$ is $x^0$ . You do not know in advance which $k$ produces it, so let the exponent tell you. With $A = x^2$ and $B = \dfrac{3}{x} = 3x^{-1}$ and $n = 6$ , the general term is

T_k = \binom{6}{k} (x^2)^{6-k} \left(3x^{-1}\right)^{k} = \binom{6}{k}\, 3^{k}\, x^{12 - 2k}\, x^{-k}.

Combine the two powers of $x$ by adding exponents, which is the whole reason for writing $\dfrac{3}{x}$ as $3x^{-1}$ :

T_k = \binom{6}{k}\, 3^{k}\, x^{12 - 3k}.

The term is constant when its exponent is zero:

12 - 3k = 0 \quad\Longrightarrow\quad k = 4.

Only $k = 4$ works, so there is exactly one constant term. Evaluate it with $\binom{6}{4} = 15$ from row $6$ and $3^4 = 81$ :

T_4 = 15 \cdot 81 = 1215.

The constant term is $1215$ . Nothing was expanded: one equation in $k$ replaced all seven terms.

A bit of history (optional)

A line of Sanskrit verse is built from syllables of two kinds, long and short. Poets wanted to know how much freedom that gave them. How many lines of a given length can the two kinds of syllable produce between them?

Pingala, a scholar in India, composed the surviving treatise on that counting question around the second century BCE. Count the lines that use exactly $k$ short syllables in $n$ positions. What you are choosing is which positions receive a short syllable, the same act you have performed all lesson, choosing which factors donate their $y$ . A single number therefore answers both questions: the one you have been writing as $\binom{n}{k}$ .

Pingala’s own verses are terse, and scholars date the first surviving Pascal-style construction of the triangle to a later commentator, Virahanka, in the seventh century CE. A tenth-century commentary by Halayudha spells the construction out explicitly and names the resulting figure the meru-prastara, the staircase of Mount Meru. Every row is built from the row above it by addition, which is Pascal’s rule operating with no algebra anywhere near it, centuries before the triangle carried Pascal’s name in the West.