D1 - Binomial expansions for integer powers | AQA A-Level Mathematics

Learn how binomial coefficients count choices in an expansion. You will expand positive integer powers, select individual coefficients and explain why the same counting appears in binomial probabilities.

Factorials and binomial coefficients

Six lamps each have an on/off switch. How many settings have exactly three lamps on? There are $2^6=64$ settings altogether, but simply halving that number does not answer the question. We need to count choices of positions.

A binomial has two terms. Start with three brackets: $(a+b)^3=(a+b)(a+b)(a+b)$ . Choosing $b$ from just one bracket makes $a^2b$ . There are three possible positions for that choice, so the coefficient is $3$ . The expansion records choices, not just powers.

You already know

Multiplying brackets means choosing one term from each bracket and adding the resulting products. Laws of indices give $a^2a=a^3$ .

For larger counts, factorial notation keeps the multiplication manageable.

Factorial

For a positive integer $n$ , $n!=n(n-1)\cdots2\times1$ , and $0!=1$ . For example, $4!=24$ , whereas $4!\ne4^4$ . The factorial definition is not printed separately in the formula booklet: learn it.

Choosing two positions from seven gives $7\times6$ ordered choices, but each pair has been counted twice. Divide by $2!$ to forget the order. In general, this gives the binomial coefficient, written $\binom nr$ , ${}^nC_r$ or $nCr$ .

Binomial coefficient

\binom nr={}^nC_r=\frac{n!}{r!(n-r)!},\qquad 0\le r\le n

Here $n,r\in\mathbb Z_0^+=\{0,1,2,\ldots\}$ , the non-negative integers, with $r\le n$ . In particular, $\binom n0=\binom nn=1$ : there is one way to choose nothing and one way to choose everything.
In the formula booklet.

Cancel common factorial factors before evaluating. This reveals the count and avoids unnecessarily large numbers.

Worked example 1: choosing two positions

Question. Calculate $\binom72$ and explain why $\binom75$ has the same value.

Plan. Cancel $5!$ , then interpret the complementary choice.

\begin{aligned} \binom72&=\frac{7!}{2!5!} && \text{use the coefficient formula}\\ &=\frac{7\times6}{2\times1}=21 && \text{cancel }5! \end{aligned}

Why this step? Choosing the two excluded positions determines the five included positions, so each choice has exactly one partner.

Answer. $\binom72=\binom75=21$ .

Check. The factorial denominators $2!5!$ and $5!2!$ are equal.

The next example changes only the number chosen from two to three. There are now $3!$ orders of each selection to remove.

Worked example 2: choosing three positions

Question. Calculate $\binom73$ .

Plan. Keep the three uncancelled numerator factors and divide by $3!$ .

\begin{aligned} \binom73&=\frac{7!}{3!4!} && \text{choose three positions}\\ &=\frac{7\times6\times5}{3\times2\times1}=35 && \text{cancel and divide} \end{aligned}

Why this step? Dividing only by $3$ would leave each unordered selection counted twice; all $3!=6$ orders must be removed.

Answer. $\binom73=35$ .

Check. Complementary choices give $\binom74=35$ too.

The symmetry $\binom nr=\binom n{n-r}$ follows from that complementary-choice argument. It is not printed separately in the formula booklet; derive it from the supplied factorial formula.

Watch out

Wrong: $\binom73=7\times6\times5/3$ . This removes too few orders. Right: divide by $3!=6$ , giving $35$ .

Now increase the number of available positions, while keeping three selections.

Expanding a positive integer power

In $(a+b)^3$ , choosing the second term zero, one, two or three times gives coefficients $1,3,3,1$ . With a positive integer power, every choice is counted and the expansion ends.

Positive integer binomial expansion

For $n\in\mathbb N$ ,

(a+b)^n=a^n+\binom n1a^{n-1}b+\binom n2a^{n-2}b^2+\cdots+b^n

Replace $b$ by the whole second term, such as $3x$ or $-3x$ . This is an exact identity for every real input for which the bracket is defined.
In the formula booklet.

The first-term power falls while the second-term power rises; their exponents add to $n$ . There are $n+1$ terms before any zero coefficients are removed. That count follows from the supplied expansion; it is not a separately supplied formula.

Worked example 3: a complete expansion

Question. Expand $(2+3x)^4$ in ascending powers of $x$ .

Plan. Use coefficients $1,4,6,4,1$ , retaining the powers of both bracket terms.

\begin{aligned} (2+3x)^4&=2^4+4(2^3)(3x)+6(2^2)(3x)^2\\ &\quad+4(2)(3x)^3+(3x)^4 && \text{powers total four}\\ &=16+96x+216x^2+216x^3+81x^4 && \text{evaluate each factor} \end{aligned}

Why this step? The coefficient $6$ multiplies $2^2(3x)^2$ ; it is not itself the coefficient of $x^2$ .

Answer. $16+96x+216x^2+216x^3+81x^4$ .

Check. At $x=1$ , the coefficients add to $625=5^4$ ; at $x=0$ , both forms give $16$ .

Change only the sign of the second bracket term. The odd powers change sign; the even powers do not.

Worked example 4: keeping a negative term intact

Question. Expand $(2-3x)^4$ .

Plan. Use the same coefficients, with the whole second term $-3x$ inside brackets.

\begin{aligned} (2-3x)^4&=2^4+4(2^3)(-3x)+6(2^2)(-3x)^2\\ &\quad+4(2)(-3x)^3+(-3x)^4 && \text{retain the signed term}\\ &=16-96x+216x^2-216x^3+81x^4 && \text{odd powers are negative} \end{aligned}

Why this step? $(-3x)^2=9x^2$ , while $(-3x)^3=-27x^3$ . Brackets protect both the sign and the factor $3$ .

Answer. $16-96x+216x^2-216x^3+81x^4$ .

Check. At $x=1$ , the original gives $1$ and the expansion gives $16-96+216-216+81=1$ .

Substitution checks can expose an error, although the expansion method establishes the identity for every input.

Watch out

Wrong: $(2-3x)^4=16+81x^4$ . This loses all the mixed choices. Right: include all five terms, including the negative odd-power terms.

Keep the same power and negative sign, but change the size of the second term.

Finding selected terms and coefficients

If you need just the coefficient of $x^3$ , most of a full expansion is unnecessary. In $(3+2x)^5$ , exactly three selections of $2x$ produce that power.

Selecting a term

The term formed by choosing $bx$ exactly $r$ times is

\binom nr a^{n-r}(bx)^r

It is term number $r+1$ because the first term has $r=0$ . This general term is part of the supplied positive integer expansion.
In the formula booklet.

A coefficient is the multiplier of a specified power; the term includes that power. For a product of expansions, list every pair of powers whose exponents add to the requested exponent.

Worked example 5: one selected coefficient

Question. Find the coefficient of $x^3$ in $(3+2x)^5$ .

Plan. Select $r=3$ , then remove $x^3$ from the resulting term.

\begin{aligned} T&=\binom53 3^{5-3}(2x)^3 && \text{choose three }2x\text{ terms}\\ &=10\times9\times8x^3=720x^3 && \text{evaluate the factors} \end{aligned}

Why this step? Choosing $r=2$ finds the third term, which contains $x^2$ , rather than the requested $x^3$ .

Answer. The coefficient is $720$ .

Check. Direct choices give $10$ arrangements, each contributing $3^2\times2^3=72$ to the coefficient.

The next example adds only a linear factor. That creates a second route to the requested power.

Worked example 6: a coefficient in a product

Question. Find the coefficient of $x^3$ in $(1-x)(3+2x)^5$ .

Plan. Combine the cubic term times $1$ and the quadratic term times $-x$ .

\begin{aligned} [x^2]\,(3+2x)^5&=\binom52 3^3 2^2=1080 && \text{select two }2x\text{ terms}\\ [x^3]\,(1-x)(3+2x)^5&=720-1080=-360 && \text{add both contributions} \end{aligned}

Here $[x^k]$ means “coefficient of $x^k$ ”; this shorthand is not supplied in the formula booklet and need not be used in your answer.

Why this step? The products $1\times x^3$ and $(-x)\times x^2$ both contain $x^3$ , so both contribute.

Answer. The coefficient is $-360$ .

Check. The relevant product is $(1-x)(\cdots+1080x^2+720x^3+\cdots)$ , whose cubic terms are $720x^3-1080x^3$ .

A reciprocal changes the exponent matching, not the counting. In $(2-x^{-1})^4$ , the general term has power $x^{-r}$ . The coefficient of $x^{-2}$ is therefore $\binom42 2^2(-1)^2=24$ , valid for $x\ne0$ . In a product involving positive and negative powers, a constant term comes from exponent pairs adding to zero.

Watch out

Wrong: multiply only the two cubic coefficients of two factors. Right: add the products from every exponent pair adding to $3$ . A constant factor can contribute too.

Change only the linear factor in the matched problem.

Why binomial probabilities use the same coefficients

Four tests with exactly two passes could have outcomes pass, pass, fail, fail, or any other arrangement of those two passes. Counting their positions is the same choice as selecting two second terms from four brackets.

For independent trials with a constant success probability $p$ , one particular arrangement of $r$ successes has probability $p^r(1-p)^{n-r}$ . There are $\binom nr$ such arrangements. They are mutually exclusive, so their probabilities add.

Binomial probability

\mathrm P(\text{exactly }r\text{ successes})=\binom nr p^r(1-p)^{n-r}

This requires a fixed integer number $n$ of trials, two outcomes per trial, independence and the same success probability $p$ on every trial, with $0\le p\le1$ .
In the formula booklet.

The coefficient counts arrangements; it is not a probability itself. AQA permits calculator probability functions. Here, writing the factors reveals why the formula works; later Statistics lessons develop model choice and distribution calculations.

Worked example 7: exactly two passes

Question. A device independently passes each of four tests with probability $0.6$ . Find the probability of exactly two passes.

Plan. Multiply one arrangement's probability by the number of choices of successful tests.

\begin{aligned} N&=\binom42=6 && \text{choose the passed tests}\\ \mathrm P(2\text{ passes})&=6(0.6)^2(0.4)^2 && \text{add six equal probabilities}\\ &=0.3456 && \text{evaluate} \end{aligned}

Why this step? Independence allows multiplication within an arrangement; mutual exclusivity allows addition across arrangements.

Answer. $0.3456$ .

Check. One arrangement has probability $0.0576$ , so six arrangements give $0.3456<1$ .

Change only the requested number of passes. Both the count of arrangements and the exponents must change.

Worked example 8: exactly three passes

Question. For the same four independent tests, each passed with probability $0.6$ , find the probability of exactly three passes.

Plan. Choose three successful positions and use three pass factors and one fail factor.

\begin{aligned} N&=\binom43=4 && \text{choose three positions}\\ \mathrm P(3\text{ passes})&=4(0.6)^3(0.4) && \text{three passes, one failure}\\ &=0.3456 && \text{evaluate} \end{aligned}

Why this step? “Exactly three” includes the failure factor; $(0.6)^3$ alone does not specify what happens in the fourth test.

Answer. $0.3456$ , coincidentally the same probability as exactly two passes.

Check. The ratio of the two probabilities is $(4/6)(0.6/0.4)=1$ .

All possible success counts correspond to the terms of $((1-p)+p)^n$ , so their probabilities add to $1$ . For the six switches, the counts are $1,6,15,20,15,6,1$ ; they add to $64$ . Exactly three on occurs in $\binom63=20$ settings. A central count can be the largest without being half the total.

Watch out

Wrong: $(0.6)^3$ for exactly three passes in four tests. Right: $4(0.6)^3(0.4)$ , including the unsuccessful trial and all four possible positions for it.

Now use a different trial count and success probability, keeping the same counting method.

Quick Recap

Factorials count orders; dividing by $r!$ removes the orders within a selection.
Keep both bracket terms, their signs and their powers in a binomial expansion.
Select only the powers that can contribute to a requested coefficient.
The same coefficient counts the mutually exclusive arrangements in a binomial probability.

The opening question has $\binom63=20$ settings with exactly three lamps on, out of $64$ settings altogether.

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D1 - Binomial expansions for integer powers

Factorials and binomial coefficients

Expanding a positive integer power

Finding selected terms and coefficients

Why binomial probabilities use the same coefficients

Quick Recap

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