The Binomial Theorem from Subset Counting

Why Subset Counting Produces the Binomial Theorem

A power such as \((a+b)^n\) is a product of \(n\) identical factors. When that product is expanded, each resulting term comes from choosing either \(a\) or \(b\) from every factor. The key question is not just what the resulting powers are, but how many different choices produce each term. Subset counting answers that question and explains the coefficients in the binomial theorem.

This connection is useful in algebraic expansion and in probability calculations that involve counting selections. For an exam problem, the reliable method is to identify how many factors contribute the second term, then use that number consistently for the coefficient and the powers.

Definition: For a nonnegative integer \(n\) and an integer \(k\) with \(0\leq k\leq n\), the binomial coefficient \(\binom{n}{k}\) is the number of ways to choose a subset of size \(k\) from a set of \(n\) distinct positions. Equivalently, \(\binom{n}{k}=\frac{n!}{k!(n-k)!}\).

The Binomial Theorem

Theorem: If \(n\) is a nonnegative integer and \(a\) and \(b\) are commuting quantities, then

(a+b)^n=\sum_{k=0}^{n}\binom{n}{k}a^{n-k}b^k =\binom{n}{0}a^n+\binom{n}{1}a^{n-1}b+\cdots+\binom{n}{n}b^n.

The index \(k\) counts how many of the \(n\) factors contribute \(b\), the second term. The remaining \(n-k\) factors contribute \(a\). There are \(\binom{n}{k}\) ways to choose those \(k\) positions, and every such choice produces the same product \(a^{n-k}b^k\).

The condition that \(n\) is a nonnegative integer matters: the proof below uses a product of \(n\) factors. This finite expansion does not, by itself, give a formula for a non-integer exponent.

Derivation by Grouping Factor Choices

Write the power as a product with labelled positions. A position means one particular factor in that product, even though all factors have the same form. Distributing the product creates one term for every pattern of choices.

Fix the number of second-term choices.
Suppose exactly \(k\) of the \(n\) factor positions contribute \(b\). Then the other \(n-k\) positions contribute \(a\).

Count the patterns.
Choosing which \(k\) positions contribute \(b\) is choosing a \(k\)-element subset of \(n\) positions. There are \(\binom{n}{k}\) such subsets.

Write the product for each pattern.
Every pattern with \(k\) second-term choices contributes \(a^{n-k}b^k\). Since there are \(\binom{n}{k}\) patterns, their combined contribution is \(\binom{n}{k}a^{n-k}b^k\).

Group all possible counts.
The number \(k\) can be any integer from \(0\) to \(n\). Adding the contribution for each value of \(k\) gives the theorem.

This argument also explains why the coefficient counts choices rather than merely recording a numerical property of the symbols \(a\) and \(b\). For instance, when \(k=2\), every pair of factor positions gives one occurrence of \(a^{n-2}b^2\). The number of such pairs is \(\binom{n}{2}\).

Worked Example: Expanding a Binomial with Positive Terms

Expand \((x+2y)^5\). For each term, state how many factors contribute \(2y\), and verify the expansion by setting \(x=y=1\).

Method. Apply the binomial theorem with \(a=x\), \(b=2y\), and \(n=5\). If exactly \(k\) factors contribute \(2y\), the term is \(\binom{5}{k}x^{5-k}(2y)^k\).

For \(k=0\), the term is \(\binom{5}{0}x^5(2y)^0=x^5\).

For \(k=1\), it is \(\binom{5}{1}x^4(2y)=5\cdot2x^4y=10x^4y\).

For \(k=2\), it is \(\binom{5}{2}x^3(2y)^2=10\cdot4x^3y^2=40x^3y^2\).

For \(k=3\), it is \(\binom{5}{3}x^2(2y)^3=10\cdot8x^2y^3=80x^2y^3\).

For \(k=4\), it is \(\binom{5}{4}x(2y)^4=5\cdot16xy^4=80xy^4\).

For \(k=5\), it is \(\binom{5}{5}(2y)^5=32y^5\). Therefore,

$$ (x+2y)^5=x^5+10x^4y+40x^3y^2+80x^2y^3+80xy^4+32y^5. $$

Check. Setting \(x=y=1\), the left side is \(3^5=243\). The sum of the coefficients on the right is \(1+10+40+80+80+32=243\). The two evaluations agree. This check is useful for detecting an arithmetic error, although it does not replace checking each term.

Common mistake avoided: The coefficient for \(k=3\) is \(\binom{5}{3}\), not \(\binom{5}{1}\) simply because the term also has two factors of \(x\). The index counts choices of the second term.

Using the Theorem to Extract a Coefficient

To find one coefficient, a full expansion is usually unnecessary. Match the requested powers to the general term. Once the matching value of \(k\) is determined, calculate the coefficient, including any sign and numerical factors inside the second term.

Worked Example: Tracking a Negative Second Term

Find the coefficient of \(u^2v^4\) in \((3u-2v)^6\). Identify how many factor positions contribute \(-2v\), and account for its sign.

Method. Write the general term with \(a=3u\), \(b=-2v\), and \(n=6\):

\binom{6}{k}(3u)^{6-k}(-2v)^k.

The power of \(v\) is \(k\). To obtain \(v^4\), set \(k=4\), so four of the six positions contribute \(-2v\). The other two contribute \(3u\), giving \(u^{6-4}=u^2\), as required.

The coefficient is therefore

\binom{6}{4}3^2(-2)^4 =15\cdot9\cdot16 =2160.

The sign is positive because \((-2)^4\) is positive. The requested coefficient is \(2160\).

Check. The term itself is \(2160u^2v^4\), with the requested exponents. The method avoids confusing the number of first-term choices with the number of second-term choices.

Worked Example: Matching Exponents Before Calculating

Find the term containing \(x^6\) in \((3x^2+y)^5\), and give its coefficient.

Method. If \(k\) factors contribute \(y\), then \(5-k\) factors contribute \(3x^2\). The power of \(x\) in that term is \(2(5-k)\). Match it to the requested exponent:

2(5-k)=6,\qquad 5-k=3,\qquad k=2.

Thus two factors contribute \(y\), and three contribute \(3x^2\). The term is

\binom{5}{2}(3x^2)^3y^2 =10\cdot27x^6y^2 =270x^6y^2.

The coefficient of \(x^6y^2\) is \(270\). Matching the exponent first is important: the index \(k\) is determined by the power of \(x\), not by the power of \(3x^2\) alone.

Check. Three factors of \(3x^2\) produce \(x^{2\cdot3}=x^6\), while the other two factors produce \(y^2\). The term has the required power of \(x\).

Coefficients, Signs, and Useful Checks

For a binomial \((a+b)^n\), the coefficient \(\binom{n}{k}\) is always nonnegative. A negative coefficient in an expansion comes from a negative value within a term, such as \((-c)^k\), not from the subset count. If \(b=-c\), then the \(k\)th grouped contribution is \(\binom{n}{k}a^{n-k}(-c)^k\). Its sign is positive when \(k\) is even and negative when \(k\) is odd, provided \(a\) and \(c\) are positive.

A general coefficient calculation should keep three pieces separate until the end:

Subset count: \(\binom{n}{k}\), where \(k\) is the number of second-term positions.
Numerical factors: the constants raised to their respective powers.
Variable powers and sign: the powers follow from the same \(k\), and a negative second term contributes a sign determined by \(k\).

The subset-count interpretation also gives simple checks on the coefficients. Setting \(a=b=1\) in the theorem shows that the sum of the coefficients is \(2^n\), because there are two choices at each of \(n\) positions. Setting \(a=1\) and \(b=-1\) gives an alternating sum of zero for \(n>0\), since \((1-1)^n=0\). These checks can reveal an incorrect expansion, but they cannot guarantee that every individual term is correct.

Common Mistake: Using the Wrong Exponent on a Constant

Consider the coefficient of \(r^2s^3\) in \((2r-s)^5\). To get \(s^3\), exactly three positions must contribute \(-s\). The other two positions contribute \(2r\). A tempting but incorrect calculation is

\binom{5}{3}2^3(-1)^3=-80.

This gives the constant \(2\) an exponent of three, even though \(2\) occurs only in the two first-term choices. The corrected calculation is

\binom{5}{3}2^2(-1)^3 =10\cdot4\cdot(-1) =-40.

The coefficient is \(-40\). As a check on the factor counts, three choices of \(-s\) give \(s^3\), and the remaining two choices of \(2r\) give \(r^2\) and the numerical factor \(2^2\). The binomial coefficient counts which three positions supply the second term; it does not determine which constant receives which exponent.

Key takeaway: In the \(k\)th term of \((a+b)^n\), \(k\) positions contribute the second term and \(n-k\) positions contribute the first. The coefficient \(\binom{n}{k}\) counts the possible position sets. Determine \(k\) from the requested powers before calculating constants or signs.

Check Your Understanding

For each problem, identify the relevant second-term count before calculating. Answers are included for checking.

Find the coefficient of \(x^2y^2\) in \((x+3y)^4\). (Answer: \(54\))
Find the coefficient of \(a^3b^3\) in \((2a-b)^6\), including its sign. (Answer: \(-160\))
Find the coefficient of \(t^4\) in \((1+2t)^5\). State how many factor positions contribute \(2t\). (Answer: \(80\); four positions)
Find the term containing \(x^6\) in \((3x^2+y)^5\). (Answer: \(270x^6y^2\))
In \((2p-q)^7\), find the coefficient of \(p^4q^3\). State how many positions contribute \(-q\) and verify the sign. (Answer: \(-560\))

What You'll Learn