7.7 Binomial Theorem
Learning Objectives
By the end of this section, you will be able to:
- Recognize the pattern of coefficients and powers in a binomial expansion \((x + y)^n\).
- Use the Binomial Theorem to expand \((x + y)^n\) without multiplying out the factors.
- Find a specific term (or the coefficient of a specific term) inside a binomial expansion.
We close the chapter with one more payoff from combinations. Multiplying \((x + y)^n\) out by hand gets miserable fast — by the time you reach \((x + y)^{10}\) there are over a thousand terms to collect. The Binomial Theorem replaces that grind with a tidy formula whose coefficients are exactly the combination counts \(\binom{n}{r}\) you already know how to compute.
7.7.1 Patterns in Binomial Expansions
Before we state the theorem, let's look at what the first several expansions actually do. Multiply each by hand and you get:
$$(x + y)^2 = x^2 + 2xy + y^2$$ $$(x + y)^3 = x^3 + 3x^2 y + 3xy^2 + y^3$$ $$(x + y)^4 = x^4 + 4x^3 y + 6x^2 y^2 + 4xy^3 + y^4$$ $$(x + y)^5 = x^5 + 5x^4 y + 10x^3 y^2 + 10x^2 y^3 + 5xy^4 + y^5$$ $$(x + y)^6 = x^6 + 6x^5 y + 15x^4 y^2 + 20x^3 y^3 + 15x^2 y^4 + 6xy^5 + y^6$$Three patterns jump out.
- The expansion of \((x + y)^n\) has exactly \(n + 1\) terms.
- In each term, the powers of \(x\) and \(y\) add to \(n\).
- The powers of \(x\) start at \(n\) and step down by one each term; the powers of \(y\) start at \(0\) and step up by one each term.
So the shape of the expansion is easy — the only question left is what are the coefficients?
The expansion \((x + y)^3 = (x + y)(x + y)(x + y)\) becomes a product of three factors. If we multiply out without collecting, every product is built by picking an \(x\) or a \(y\) from each factor:
$$xxx + xxy + xyx + xyy + yxx + yxy + yyx + yyy.$$The term \(x^2 y\) happens exactly when we pick \(x\) from two of the three factors and \(y\) from the third. The number of ways to do that is \(\binom{3}{2} = 3\). That is why the coefficient of \(x^2 y\) in \((x + y)^3\) is \(3\) — it is literally counting how many of the eight products simplify to \(x^2 y\).
Every coefficient in a binomial expansion is a combination count. To read off the coefficient of \(x^{n-r} y^r\) in \((x + y)^n\), ask: out of \(n\) factors, in how many ways can I pick the \(r\) that contribute a \(y\)? The answer is \(\binom{n}{r}\). The powers of \(x\) and \(y\) track the choice; the coefficient counts the ways to make the choice.
Source: Applied Finite Mathematics
The binomial coefficient \(\binom{n}{r}\), also written \(_n C_r\), is the number of ways to choose \(r\) objects from \(n\):
$$\binom{n}{r} = {}_n C_r = \dfrac{n!}{r!\,(n - r)!}.$$In a binomial expansion, \(\binom{n}{r}\) is the coefficient of the term \(x^{n-r} y^r\).
Source: Applied Finite Mathematics
Find the coefficient of the term \(x^2 y^5\) in the expansion of \((x + y)^7\).
Solution
Step 1 — Rewrite the seventh power as seven factors.
$$(x + y)^7 = (x + y)(x + y)(x + y)(x + y)(x + y)(x + y)(x + y).$$Step 2 — Decide how to get \(x^2 y^5\) from the product. We need an \(x\) from exactly two of the seven factors and a \(y\) from the other five.
Step 3 — Count the ways. Choose which \(2\) of the \(7\) factors contribute an \(x\):
$$\binom{7}{2} = \dfrac{7!}{2!\,5!} = 21.$$Answer: The coefficient of \(x^2 y^5\) is \(21\).
Source: Applied Finite Mathematics
Find the coefficient of the term \(x^3 y^2\) in the expansion of \((x + y)^5\).
Solution
Choose which \(2\) of the \(5\) factors contribute a \(y\):
$$\binom{5}{2} = \dfrac{5!}{2!\,3!} = 10.$$The coefficient is \(10\).
7.7.2 The Binomial Theorem
Once you believe the coefficient of \(x^{n-r} y^r\) is always \(\binom{n}{r}\), the formula for the whole expansion follows. Writing out \((x + y)^n\) term by term:
$$(x + y)^n = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} y + \binom{n}{2} x^{n-2} y^2 + \cdots + \binom{n}{n-1} x y^{n-1} + \binom{n}{n} y^n.$$That is the Binomial Theorem. Every coefficient is a combination count; every exponent pair sums to \(n\); there are \(n + 1\) terms.
Expanding \((x + y)^{10}\) by repeated distribution means tracking \(2^{10} = 1{,}024\) products and then collecting like terms. The Binomial Theorem skips that entirely — it tells you there are \(11\) terms and it tells you the coefficient of each by computing a single combination. The formula is not a trick; it is the accounting that distribution would have done for you, already tidied up.
Because \(\binom{n}{r} = \binom{n}{n - r}\), the coefficients of a binomial expansion read the same forwards and backwards: \(1, n, \ldots, n, 1\). If you ever compute \(\binom{7}{5}\), you can use \(\binom{7}{2} = 21\) instead — same number, fewer factorials. That symmetry is also why Pascal's triangle is left-right mirrored.
Source: Applied Finite Mathematics
For any positive integer \(n\),
$$(x + y)^n = \sum_{r = 0}^{n} \binom{n}{r} x^{n-r} y^r = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} y + \cdots + \binom{n}{n} y^n.$$Source: Applied Finite Mathematics
Expand \((x + y)^7\).
Solution
Step 1 — Write the skeleton without coefficients. Powers of \(x\) step down from \(7\); powers of \(y\) step up from \(0\):
$$(x + y)^7 = \square\,x^7 + \square\,x^6 y + \square\,x^5 y^2 + \square\,x^4 y^3 + \square\,x^3 y^4 + \square\,x^2 y^5 + \square\,x y^6 + \square\,y^7.$$Step 2 — Fill in each coefficient with \(\binom{7}{r}\).
| Term | \(\binom{7}{r}\) | Value |
|---|---|---|
| \(x^7\) | \(\binom{7}{0}\) | \(1\) |
| \(x^6 y\) | \(\binom{7}{1}\) | \(7\) |
| \(x^5 y^2\) | \(\binom{7}{2}\) | \(21\) |
| \(x^4 y^3\) | \(\binom{7}{3}\) | \(35\) |
| \(x^3 y^4\) | \(\binom{7}{4}\) | \(35\) |
| \(x^2 y^5\) | \(\binom{7}{5}\) | \(21\) |
| \(x y^6\) | \(\binom{7}{6}\) | \(7\) |
| \(y^7\) | \(\binom{7}{7}\) | \(1\) |
Step 3 — Assemble the expansion.
$$(x + y)^7 = x^7 + 7x^6 y + 21 x^5 y^2 + 35 x^4 y^3 + 35 x^3 y^4 + 21 x^2 y^5 + 7 x y^6 + y^7.$$Answer: \((x + y)^7 = x^7 + 7x^6 y + 21 x^5 y^2 + 35 x^4 y^3 + 35 x^3 y^4 + 21 x^2 y^5 + 7 x y^6 + y^7\).
Source: Applied Finite Mathematics
Expand \((3a - 2b)^4\).
Solution
Step 1 — Match the template \((x + y)^n\). Let \(x = 3a\) and \(y = -2b\); then \(n = 4\).
Step 2 — Apply the Binomial Theorem.
$$(3a - 2b)^4 = \binom{4}{0}(3a)^4 + \binom{4}{1}(3a)^3(-2b) + \binom{4}{2}(3a)^2(-2b)^2 + \binom{4}{3}(3a)(-2b)^3 + \binom{4}{4}(-2b)^4.$$Step 3 — Evaluate each coefficient and power carefully. The coefficients are \(\binom{4}{0}, \binom{4}{1}, \binom{4}{2}, \binom{4}{3}, \binom{4}{4} = 1, 4, 6, 4, 1\). Compute each piece:
- \(1 \cdot (3a)^4 = 1 \cdot 81 a^4 = 81 a^4\) - \(4 \cdot (3a)^3 \cdot (-2b) = 4 \cdot 27 a^3 \cdot (-2 b) = -216 a^3 b\) - \(6 \cdot (3a)^2 \cdot (-2b)^2 = 6 \cdot 9 a^2 \cdot 4 b^2 = 216 a^2 b^2\) - \(4 \cdot (3a) \cdot (-2b)^3 = 4 \cdot 3 a \cdot (-8 b^3) = -96 a b^3\) - \(1 \cdot (-2b)^4 = 16 b^4\)
Step 4 — Combine.
$$(3a - 2b)^4 = 81 a^4 - 216 a^3 b + 216 a^2 b^2 - 96 a b^3 + 16 b^4.$$Answer: \((3a - 2b)^4 = 81 a^4 - 216 a^3 b + 216 a^2 b^2 - 96 a b^3 + 16 b^4\).
Source: Applied Finite Mathematics
Expand \((x - 2y)^4\).
Solution
Let \(X = x\) and \(Y = -2y\) in \((X + Y)^4\). The coefficients are \(1, 4, 6, 4, 1\):
$$(x - 2y)^4 = x^4 + 4 x^3(-2y) + 6 x^2(-2y)^2 + 4 x(-2y)^3 + (-2y)^4.$$Simplify each term:
$$(x - 2y)^4 = x^4 - 8 x^3 y + 24 x^2 y^2 - 32 x y^3 + 16 y^4.$$7.7.3 Finding a Specific Term
Sometimes you do not need the whole expansion — you just need one term, say the fifth term of \((3a - 2b)^7\). The Binomial Theorem lets you jump straight to it.
Writing the expansion in order, the \(r\)-th term uses \(\binom{n}{r - 1}\) and carries \(y\) to the power \(r - 1\):
$$\text{\(r\)-th term} = \binom{n}{r - 1}\, x^{n - (r - 1)}\, y^{r - 1}.$$The off-by-one is the one thing to watch. If you want the \(5\)-th term, the exponent on \(y\) is \(4\), not \(5\).
Textbooks number the terms starting at \(r = 1\) (the pure \(x^n\) term). Because the exponent on \(y\) starts at \(0\), it will always be one less than the term number. That's the source of every "off-by-one" mistake students make here. Write down the term number, subtract one, then plug into the formula.
Finding the fifth term of \((x + y)^{12}\) takes exactly one combination calculation — \(\binom{12}{4}\) — not twelve. The Binomial Theorem turns "find a particular term" into a lookup, which is why this technique shows up often on finite-math exams.
Source: Applied Finite Mathematics
In the expansion of \((x + y)^n\), the \(r\)-th term is
$$\binom{n}{r - 1}\, x^{n - (r - 1)}\, y^{r - 1}.$$Source: Applied Finite Mathematics
Find the fifth term of the expansion \((3a - 2b)^7\).
Solution
Step 1 — Identify the pieces. Match \((x + y)^n\) with \(x = 3a\), \(y = -2b\), \(n = 7\). The term number is \(r = 5\), so the exponent on \(y\) is \(r - 1 = 4\) and the exponent on \(x\) is \(7 - 4 = 3\).
Step 2 — Coefficient. The coefficient is \(\binom{7}{4}\):
$$\binom{7}{4} = \dfrac{7!}{4!\,3!} = 35.$$Step 3 — Assemble the term.
$$(35)(3a)^3(-2b)^4 = 35 \cdot 27 a^3 \cdot 16 b^4.$$Step 4 — Multiply the constants. \(35 \cdot 27 = 945\) and \(945 \cdot 16 = 15{,}120\).
Answer: The fifth term is \(15{,}120\, a^3 b^4\).
Source: Applied Finite Mathematics
Find the third term of the expansion \((2x - 3y)^6\).
Solution
Match \((x + y)^n\) with \(x = 2x\), \(y = -3y\), \(n = 6\). For the third term, \(r = 3\) so the exponent on \(y\) is \(2\) and on \(x\) is \(4\). The coefficient is
$$\binom{6}{2} = \dfrac{6!}{2!\,4!} = 15.$$The third term is
$$15\,(2x)^4(-3y)^2 = 15 \cdot 16 x^4 \cdot 9 y^2 = 2{,}160\, x^4 y^2.$$Summary
Patterns in \((x + y)^n\)
- \(n + 1\) terms.
- In each term, the exponents of \(x\) and \(y\) sum to \(n\).
- Powers of \(x\) start at \(n\) and decrease; powers of \(y\) start at \(0\) and increase.
Binomial Theorem
For any positive integer \(n\),
$$(x + y)^n = \sum_{r = 0}^{n} \binom{n}{r} x^{n-r} y^r = \binom{n}{0} x^n + \binom{n}{1} x^{n-1} y + \cdots + \binom{n}{n} y^n.$$
\(r\)-th Term Formula
The \(r\)-th term of \((x + y)^n\) is
$$\binom{n}{r - 1}\, x^{n - (r - 1)}\, y^{r - 1}.$$Remember: the exponent on \(y\) is one less than the term number.
Problem Set 7.7
Source: Applied Finite Mathematics
Use the Binomial Theorem to do the following problems.
Problem 1. Expand \((a + b)^5\).
Problem 1 Solution
Step 1 — Identify \(n\) and the coefficients. For \((a + b)^5\), \(n = 5\). The coefficients \(\binom{5}{r}\) for \(r = 0, 1, 2, 3, 4, 5\) are \(1, 5, 10, 10, 5, 1\).
Step 2 — Apply the Binomial Theorem. Powers of \(a\) step down from \(5\); powers of \(b\) step up from \(0\):
$$(a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5.$$Answer: \((a + b)^5 = a^5 + 5 a^4 b + 10 a^3 b^2 + 10 a^2 b^3 + 5 a b^4 + b^5\).
Problem 2. Expand \((a - b)^6\).
Problem 2 Solution
Step 1 — Identify \(n\) and the coefficients. For \((a - b)^6\), \(n = 6\). The coefficients \(\binom{6}{r}\) are \(1, 6, 15, 20, 15, 6, 1\).
Step 2 — Set \(y = -b\) in the Binomial Theorem. Because \(y = -b\), each term carries a factor \((-1)^r\), so the signs alternate starting positive:
$$(a - b)^6 = a^6 - 6 a^5 b + 15 a^4 b^2 - 20 a^3 b^3 + 15 a^2 b^4 - 6 a b^5 + b^6.$$Answer: \((a - b)^6 = a^6 - 6 a^5 b + 15 a^4 b^2 - 20 a^3 b^3 + 15 a^2 b^4 - 6 a b^5 + b^6\).
Problem 3. Expand \((x - 2y)^5\).
Problem 3 Solution
Step 1 — Match the template \((x + y)^n\). Let \(y = -2y\) and \(n = 5\). Coefficients \(\binom{5}{r}\) are \(1, 5, 10, 10, 5, 1\).
Step 2 — Apply the theorem.
$$(x - 2y)^5 = x^5 + 5 x^4 (-2y) + 10 x^3 (-2y)^2 + 10 x^2 (-2y)^3 + 5 x (-2y)^4 + (-2y)^5.$$Step 3 — Simplify each term.
- \(5 x^4 (-2y) = -10 x^4 y\) - \(10 x^3 (-2y)^2 = 10 x^3 \cdot 4 y^2 = 40 x^3 y^2\) - \(10 x^2 (-2y)^3 = 10 x^2 \cdot (-8 y^3) = -80 x^2 y^3\) - \(5 x (-2y)^4 = 5 x \cdot 16 y^4 = 80 x y^4\) - \((-2y)^5 = -32 y^5\)
Step 4 — Combine.
$$(x - 2y)^5 = x^5 - 10 x^4 y + 40 x^3 y^2 - 80 x^2 y^3 + 80 x y^4 - 32 y^5.$$Answer: \((x - 2y)^5 = x^5 - 10 x^4 y + 40 x^3 y^2 - 80 x^2 y^3 + 80 x y^4 - 32 y^5\).
Problem 4. Expand \((2x - 3y)^4\).
Problem 4 Solution
Step 1 — Match the template \((x + y)^n\). Let \(x = 2x\), \(y = -3y\), \(n = 4\). Coefficients \(\binom{4}{r}\) are \(1, 4, 6, 4, 1\).
Step 2 — Apply the theorem.
$$(2x - 3y)^4 = (2x)^4 + 4 (2x)^3 (-3y) + 6 (2x)^2 (-3y)^2 + 4 (2x) (-3y)^3 + (-3y)^4.$$Step 3 — Simplify each term.
- \((2x)^4 = 16 x^4\) - \(4 (2x)^3 (-3y) = 4 \cdot 8 x^3 \cdot (-3 y) = -96 x^3 y\) - \(6 (2x)^2 (-3y)^2 = 6 \cdot 4 x^2 \cdot 9 y^2 = 216 x^2 y^2\) - \(4 (2x) (-3y)^3 = 4 \cdot 2 x \cdot (-27 y^3) = -216 x y^3\) - \((-3y)^4 = 81 y^4\)
Step 4 — Combine.
$$(2x - 3y)^4 = 16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4.$$Answer: \((2x - 3y)^4 = 16 x^4 - 96 x^3 y + 216 x^2 y^2 - 216 x y^3 + 81 y^4\).
Problem 5. Find the third term of \((2x - 3y)^6\).
Problem 5 Solution
Step 1 — Identify the pieces. Match \((x + y)^n\) with \(x = 2x\), \(y = -3y\), \(n = 6\). For the third term, \(r = 3\), so the exponent on \(y\) is \(r - 1 = 2\) and the exponent on \(x\) is \(6 - 2 = 4\).
Step 2 — Coefficient.
$$\binom{6}{2} = \dfrac{6!}{2!\,4!} = 15.$$Step 3 — Assemble the term.
$$15\,(2x)^4(-3y)^2 = 15 \cdot 16 x^4 \cdot 9 y^2.$$Step 4 — Multiply the constants. \(15 \cdot 16 = 240\) and \(240 \cdot 9 = 2{,}160\).
Answer: The third term is \(2{,}160\, x^4 y^2\).
Problem 6. Find the sixth term of \((5x + y)^8\).
Problem 6 Solution
Step 1 — Identify the pieces. Match \((x + y)^n\) with \(x = 5x\), \(y = y\), \(n = 8\). For the sixth term, \(r = 6\), so the exponent on \(y\) is \(5\) and the exponent on \(x\) is \(3\).
Step 2 — Coefficient.
$$\binom{8}{5} = \binom{8}{3} = \dfrac{8!}{3!\,5!} = 56.$$Step 3 — Assemble the term.
$$56\,(5x)^3\,y^5 = 56 \cdot 125 x^3 \cdot y^5.$$Step 4 — Multiply the constants. \(56 \cdot 125 = 7{,}000\).
Answer: The sixth term is \(7{,}000\, x^3 y^5\).
Problem 7. Find the coefficient of the \(x^3 y^4\) term in the expansion of \((2x + y)^7\).
Problem 7 Solution
Step 1 — Identify the pieces. Match \((x + y)^n\) with \(x = 2x\), \(y = y\), \(n = 7\). The target term is \(x^3 y^4\), so the exponent on \(y\) is \(4\) — meaning \(r = 4\) in \(\binom{n}{r}\).
Step 2 — Compute \(\binom{7}{4}\).
$$\binom{7}{4} = \dfrac{7!}{4!\,3!} = 35.$$Step 3 — Assemble the term and pull out the coefficient.
$$35\,(2x)^3\, y^4 = 35 \cdot 8 x^3 \cdot y^4 = 280\, x^3 y^4.$$Answer: The coefficient of \(x^3 y^4\) is \(280\).
Problem 8. Find the coefficient of the \(a^4 b^6\) term in the expansion of \((3a - b)^{10}\).
Problem 8 Solution
Step 1 — Identify the pieces. Match \((x + y)^n\) with \(x = 3a\), \(y = -b\), \(n = 10\). The target term is \(a^4 b^6\), so the exponent on the \(y\)-slot \((-b)\) is \(6\) — meaning \(r = 6\) in \(\binom{n}{r}\).
Step 2 — Compute \(\binom{10}{6}\).
$$\binom{10}{6} = \binom{10}{4} = \dfrac{10!}{4!\,6!} = 210.$$Step 3 — Assemble the term.
$$210\,(3a)^4\,(-b)^6 = 210 \cdot 81 a^4 \cdot b^6.$$Because \((-b)^6 = b^6\), the sign is positive.
Step 4 — Multiply the constants. \(210 \cdot 81 = 17{,}010\).
Answer: The coefficient of \(a^4 b^6\) is \(17{,}010\).
Problem 9. A coin is tossed 5 times. In how many ways is it possible to get three heads and two tails?
Problem 9 Solution
Step 1 — Connect coin tosses to binomial coefficients. The number of ways to get \(k\) heads in \(n\) tosses is \(\binom{n}{k}\) — the same count as the coefficient of \(H^k T^{n-k}\) in \((H + T)^n\).
Step 2 — Apply with \(n = 5\) and \(k = 3\).
$$\binom{5}{3} = \dfrac{5!}{3!\,2!} = 10.$$Answer: \(10\) ways.
Problem 10. A coin is tossed 10 times. In how many ways is it possible to get seven heads and three tails?
Problem 10 Solution
Step 1 — Apply the same reasoning as Problem 7.7.9 with \(n = 10\) and \(k = 7\).
$$\binom{10}{7} = \binom{10}{3} = \dfrac{10!}{3!\,7!} = 120.$$Answer: \(120\) ways.
Problem 11. How many subsets are there of a set that has 6 elements?
Problem 11 Solution
Step 1 — Count subsets by size. A set with \(6\) elements has \(\binom{6}{r}\) subsets of size \(r\), for \(r = 0, 1, \ldots, 6\). The total is
$$\binom{6}{0} + \binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} + \binom{6}{6}.$$Step 2 — Recognize the Binomial Theorem with \(x = y = 1\). Setting \(x = 1\) and \(y = 1\) in \((x + y)^n\) gives
$$(1 + 1)^n = \sum_{r = 0}^{n} \binom{n}{r} \cdot 1^{n-r} \cdot 1^r = \sum_{r = 0}^{n} \binom{n}{r} = 2^n.$$Step 3 — Evaluate at \(n = 6\). \(2^6 = 64\).
Answer: \(64\) subsets.
Problem 12. How many subsets are there of a set that has \(n\) elements?
Problem 12 Solution
Step 1 — Count subsets by size. A set with \(n\) elements has \(\binom{n}{r}\) subsets of each size \(r\); the total is \(\sum_{r = 0}^{n} \binom{n}{r}\).
Step 2 — Apply the Binomial Theorem with \(x = y = 1\).
$$(1 + 1)^n = \sum_{r = 0}^{n} \binom{n}{r} = 2^n.$$Answer: \(2^n\) subsets.