1.04a, 1.04b - Integer binomial expansion and combinations | OCR A-Level Mathematics A

Find a chosen coefficient without multiplying out every bracket. You will connect factorials and Pascal's triangle to integer binomial expansions, then see why the same numbers count success patterns in probability.

Factorials and binomial coefficients

Can you find the coefficient of $x^3$ in $(3+2x)^5$ without expanding the whole expression? In five copies of $(3+2x)$ , an $x^3$ term comes from choosing $2x$ from three brackets and $3$ from the other two. First we need to count those choices.

You already know

Expanding a product means choosing one term from each bracket, multiplying the chosen terms and adding the results. For example, $(a+b)^2=a^2+2ab+b^2$ : there are two ways to obtain $ab$ .

Choosing the first and second bracket gives the same selection as choosing the second and first. The order in which you name the positions must not create an extra choice.

Factorial and binomial coefficient

For a positive integer $n$ , $n!=n(n-1)\cdots2\times1$ , and $0!=1$ . The binomial coefficient $\binom nr$ , also written ${}^nC_r$ or ${}_nC_r$ , counts selections of $r$ distinct positions from $n$ , without regard to order. For example, $\binom52=10$ , not $20$ .

There are $n!/(n-r)!$ ordered selections of $r$ positions. Each unordered selection appears $r!$ times, once for each ordering, so divide by $r!$ .

Binomial coefficients

For integers $n\ge1$ and $0\le r\le n$ ,

\binom nr=\frac{n!}{r!(n-r)!},\qquad \binom n0=\binom nn=1.

Choosing the positions left out gives $\binom nr=\binom n{n-r}$ . The factorial formula and coefficient notation are printed; symmetry follows by interchanging the denominator factors.

In the formula sheet.

Cancelling common factors is usually quicker than evaluating three large factorials.

Worked example 1: choosing two bracket positions

Question. How many selections of two positions are possible in five copies of $(a+bx)$ ?

Plan. Count unordered pairs using $\binom52$ .

\begin{aligned} \binom52&=\frac{5!}{2!3!} &&\text{choose two of five}\\ &=\frac{5\times4}{2\times1}=10 &&\text{cancel }3! \end{aligned}

Why divide by $2!$ ? Each pair was counted in its two possible orders.

Answer. There are $10$ selections.

Check. Listing pairs by their smaller position gives $4+3+2+1=10$ .

Change only the number selected from two to three. Choosing three positions is equivalent to leaving two out.

Worked example 2: counting the complementary choice

Question. Find $\binom53$ and explain its connection with $\binom52$ .

Plan. Evaluate the factorial ratio, then pair each selection with its complement.

\begin{aligned} \binom53&=\frac{5!}{3!2!} &&\text{choose three of five}\\ &=\frac{5\times4}{2\times1}=10 &&\text{cancel }3! \end{aligned}

Why the same answer? Every selection of three positions leaves exactly one selection of two positions.

Answer. $\binom53=\binom52=10$ .

Check. The denominators $3!2!$ and $2!3!$ are equal.

A calculator's combinations key evaluates this same count. Its permutations key counts ordered choices instead.

Watch out

Wrong: $\binom52=5\times4=20$ . This counts every pair twice. Right: $\binom52=(5\times4)/2!=10$ . At the endpoints there is one empty selection and one selection of everything, consistent with $0!=1$ .

Try complementary selections in a larger product.

Pascal's triangle and a finite expansion

Pascal's triangle stores these position counts: row $n$ contains $\binom n0,\binom n1,\ldots,\binom nn$ . Its edges are $1$ , and each interior entry adds the two above it.

To choose $r$ positions from $n$ , either leave the last position out and choose $r$ from $n-1$ , or include it and choose $r-1$ from $n-1$ . These disjoint possibilities explain the addition rule. Thus $\binom42=\binom32+\binom31=3+3=6$ .

[DIAGRAM: asset_name: Pascal coefficients count bracket choices; caption: Pascal coefficients count bracket choices; asset_slug: h240-pascal-triangle; description: Pascal triangle from row 0 through row 6, with the two 3 entries in row 3 feeding the highlighted 6 in row 4 and the fourth-power row identified]

Each selection with $r$ copies of $bx$ contributes $a^{n-r}(bx)^r$ . The coefficient counts how many times that contribution appears.

Finite binomial expansion

For a positive integer $n$ ,

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

It is an exact polynomial identity for every real $x$ , with $n+1$ terms before any zero coefficients are removed. This is the sheet's $(a+b)^n$ formula with $b$ replaced by $bx$ .

In the formula sheet.

A neat fact is hidden in the rows: their totals double each time. Setting $a=b=1$ in the expansion gives row total $2^n$ , so row $6$ totals $64$ .

Worked example 3: a positive inner term

Question. Expand $(1+2x)^4$ in ascending powers of $x$ .

Plan. Use row $4$ , applying each power to the whole term $2x$ .

\begin{aligned} (1+2x)^4&=1+4(2x)+6(2x)^2+4(2x)^3+(2x)^4 &&\text{row }4\\ &=1+8x+24x^2+32x^3+16x^4 &&\text{simplify each power} \end{aligned}

Why keep the brackets? The power applies to $2$ as well as $x$ .

Answer. $1+8x+24x^2+32x^3+16x^4$ .

Check. At $x=1$ , the polynomial totals $81=3^4$ .

Now change only the sign of the inner term. The even powers stay positive; the odd powers change sign.

Worked example 4: a negative inner term

Question. Expand $(1-2x)^4$ in ascending powers of $x$ .

Plan. Use the same row with $-2x$ in place of $2x$ .

\begin{aligned} (1-2x)^4&=1+4(-2x)+6(-2x)^2+4(-2x)^3+(-2x)^4 &&\text{substitute the signed term}\\ &=1-8x+24x^2-32x^3+16x^4 &&\text{odd powers are negative} \end{aligned}

Why not make every coefficient negative? A negative number squared or raised to the fourth power is positive.

Answer. $1-8x+24x^2-32x^3+16x^4$ .

Check. At $x=1$ , $1-8+24-32+16=1=(-1)^4$ .

These are finite expansions. No “small $x$ ” condition belongs on them.

Watch out

Wrong: $(-2x)^2=-4x^2$ . Right: $(-2x)^2=4x^2$ . Write the signed input in brackets before raising it to a power.

Keep the fourth power and change the magnitude of the inner coefficient.

Finding a particular term or coefficient

The opening question needs just one selection count. There is no reason to calculate the other powers first.

The contribution containing $x^r$ in $(a+bx)^n$ is $\binom nr a^{n-r}b^r x^r$ . This follows directly from the supplied finite expansion. Its coefficient is $\binom nr a^{n-r}b^r$ ; the term includes $x^r$ . The constant is the first term, so the $x^r$ term is in position $r+1$ when terms are in ascending powers.

Worked example 5: the opening coefficient

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

Plan. Choose $2x$ from three brackets and $3$ from the remaining two.

\begin{aligned} \text{coefficient of }x^3&=\binom53\,3^2\,2^3 &&\text{three variable choices}\\ &=10\times9\times8=720 &&\text{evaluate the factors} \end{aligned}

Why the power $3^2$ ? Two of the five brackets still supply their constant term.

Answer. The coefficient is $720$ , and the term is $720x^3$ .

Check. There are $10$ choices, each contributing $72x^3$ .

Change only the sign of the variable term. The requested power is odd, so its coefficient changes sign.

Worked example 6: an odd power with a minus sign

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

Plan. Keep the same selection count and include the signed factor $(-2)^3$ .

\begin{aligned} \text{coefficient of }x^3&=\binom53\,3^2(-2)^3 &&\text{retain the sign}\\ &=10\times9\times(-8)=-720 &&\text{evaluate} \end{aligned}

Why is the sign negative? Three selected variable terms give three negative factors.

Answer. The coefficient is $-720$ ; $-720x^3$ is the fourth term.

Check. Replacing $x$ by $-x$ in the previous expansion reverses precisely its odd-power coefficients.

If the exponent were $6$ instead, the same requested coefficient would be $\binom63 3^3(-2)^3=-4320$ : both the selection count and the number of constant factors change. The coefficient is never just the Pascal number.

Exam tip

If OCR asks for a coefficient, identify the number multiplying the requested power. A term such as $-720x^3$ is not itself a coefficient. Write the selection expression before evaluating it so your method is visible.

Now change the requested power while keeping the bracket and exponent from example 6.

Why the same coefficients appear in probabilities

A success pattern also selects positions. For three trials, two successes can occur as SSF, SFS or FSS. These are three different outcomes even though their success counts match.

You already know

For independent events, multiply probabilities along a specified pattern. Add probabilities of mutually exclusive patterns. If success has probability $p$ , failure has probability $q=1-p$ .

With $n$ independent two-outcome trials and the same $p$ on each, a specified pattern of $r$ successes has probability $p^r q^{n-r}$ . There are $\binom nr$ possible success-position selections.

Exactly a given number of successes

For a fixed positive integer $n$ , independent two-outcome trials and constant $0<p<1$ ,

\mathrm P(\text{exactly }r\text{ successes})=\binom nr p^r(1-p)^{n-r},\quad 0\le r\le n.

Here $r$ is an integer. This counts disjoint patterns of equal probability. At $p=0$ , all trials fail with probability $1$ ; at $p=1$ , all succeed with probability $1$ . Handle these cases directly, without substituting an expression $0^0$ .

In the formula sheet.

This is the binomial probability formula you will meet again in statistics. Here the key idea is why the coefficient is needed.

Worked example 7: two successes in three trials

Question. Three independent trials each have success probability $0.3$ . Find the probability of exactly two successes.

Plan. Multiply the probability of one pattern by its number of possible positions.

\begin{aligned} \mathrm P(\mathrm{SSF})&=(0.3)^2(0.7)=0.063 &&\text{independence}\\ \mathrm P(\text{two successes})&=\binom32(0.3)^2(0.7) &&\text{three disjoint patterns}\\ &=3(0.063)=0.189 &&\text{add their probabilities} \end{aligned}

Why multiply by $3$ ? The successes need not occupy only the first two positions.

Answer. $0.189$ .

Check. SSF, SFS and FSS each contribute $0.063$ .

Change only the required success count from two to one. Keep the trials and probability unchanged.

Worked example 8: one success instead

Question. For the same three trials, find the probability of exactly one success.

Plan. Count the three possible positions of the single success.

\begin{aligned} \mathrm P(\mathrm{SFF})&=(0.3)(0.7)^2=0.147 &&\text{one pattern}\\ \mathrm P(\text{one success})&=\binom31(0.3)(0.7)^2 &&\text{choose its position}\\ &=3(0.147)=0.441 &&\text{add disjoint patterns} \end{aligned}

Why is this not $0.189$ again? The counts agree, but each pattern now contains two failure factors instead of two success factors.

Answer. $0.441$ .

Check. The probabilities for $0,1,2,3$ successes total $0.343+0.441+0.189+0.027=1$ , just as $(0.7+0.3)^3=1$ .

The opening coefficient $720$ and these probabilities use the same position-counting idea, with different factors attached to each choice.

Watch out

Wrong: probability $=\binom nr$ . A count can exceed $1$ . Right: multiply that count by the probability of each pattern, after checking independence and constant success probability.

Use the same one-success method with one more trial and a different success probability.

Quick Recap

Factorials count ordered arrangements; dividing by $r!$ gives unordered position selections.
Pascal's triangle and $\binom nr$ supply the coefficients of a finite binomial expansion.
Apply each power to the whole signed term, including its numerical factor.
Binomial probabilities count equally likely success-position patterns before adding their probabilities.

The opening coefficient is $\binom53 3^2 2^3=720$ : just one calculation, without expanding the other five terms.

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1.04a, 1.04b - Integer binomial expansion and combinations

Factorials and binomial coefficients

Pascal's triangle and a finite expansion

Finding a particular term or coefficient

Why the same coefficients appear in probabilities

Quick Recap

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