Ms1, Ms2 - Integer binomial expansion and combinations | OCR A-Level Mathematics B (MEI)

A long product can hide a very short coefficient calculation. You will use factorials and combinations to expand positive integer powers, find selected terms and explain the connection with binomial probabilities.

Factorials and binomial coefficients

Why is the coefficient of $x^2$ in $(2+x)^6$ much larger than $15$ ? In six brackets, you must select two positions for $x$ ; the other four positions each contribute $2$ . Counting the positions is only the first job.

You already know

Each term in an expanded product comes from choosing one term from every bracket. In $(a+b)^2$ , choosing $a$ then $b$ , or $b$ then $a$ , produces two contributions of $ab$ .

A selection records which positions supply the variable term. Naming the selected positions in a different order does not change the selection.

Factorial and binomial coefficient

For a positive integer $n$ , $n!=1\times2\times\cdots\times n$ , and $0!=1$ . The coefficient ${}^nC_r$ counts ways of selecting $r$ distinct objects from $n$ , without regard to order. The notation $\binom nr$ means the same count. Here we use selections of bracket positions and trial positions.

Selecting $r$ positions in order gives $n(n-1)\cdots(n-r+1)$ possibilities. Each selection has $r!$ different orderings, which must be removed.

Combinations for binomial coefficients

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

{}^nC_r=\frac{n!}{r!(n-r)!},\qquad{}^nC_0={}^nC_n=1.

The count also satisfies ${}^nC_r={}^nC_{n-r}$ , by selecting what is left out. The sheet prints the factorial ratio; the symmetry follows from it.

In the formula sheet.

Use a combinations key if you wish, but know what it counts: permutations would distinguish different orders.

Worked example 1: selecting two positions

Question. Find ${}^6C_2$ , the number of ways to choose two variable-term positions from six brackets.

Plan. Use the factorial ratio and cancel $4!$ .

\begin{aligned} {}^6C_2&=\frac{6!}{2!4!} &&\text{unordered selections}\\ &=\frac{6\times5}{2\times1}=15 &&\text{cancel common factors} \end{aligned}

Why divide by $2$ ? Each pair has been counted in two orders.

Answer. $15$ selections.

Check. Counting by the smaller selected position gives $5+4+3+2+1=15$ .

Change only the selection size from two to four. The complementary positions offer a quicker way to think.

Worked example 2: selecting the positions left out

Question. Find ${}^6C_4$ and explain why it equals ${}^6C_2$ .

Plan. Evaluate the count and identify the pair left out of each selection.

\begin{aligned} {}^6C_4&=\frac{6!}{4!2!} &&\text{choose four positions}\\ &=15={}^6C_2 &&\text{the same factorial factors} \end{aligned}

Why is the pairing exact? Each selection of four leaves precisely two positions, and each pair left out determines one selection of four.

Answer. Both counts are $15$ .

Check. There is one way to choose no positions and one way to choose all six, so the same symmetry works at the endpoints.

The empty choice contributes the constant term of an expansion; it is a genuine single choice.

Watch out

Wrong: $0!=0$ , which would make ${}^6C_0$ undefined. Right: $0!=1$ , giving ${}^6C_0=1$ . Do not substitute fractional exponents into this factorial count.

Use complementary position counts for eight brackets.

Pascal's triangle and a finite expansion

Pascal's triangle gives the coefficients before any powers of the bracket's terms are included. Each row begins and ends with $1$ ; an interior entry is the sum of the two above it.

For example, ${}^4C_2={}^3C_2+{}^3C_1=3+3=6$ . Either the fourth position is not selected, so choose two of the first three, or it is selected, so choose one of the first three. The diagram makes those two routes visible.

[DIAGRAM: asset_name: Pascal coefficients count bracket choices; caption: Pascal coefficients count bracket choices; asset_slug: h640-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]

For every chosen set of $r$ variable positions, multiply $r$ copies of $bx$ and $n-r$ copies of $a$ .

Positive integer binomial expansion

For a positive integer $n$ ,

(a+bx)^n=a^n+{}^nC_1a^{n-1}(bx)+{}^nC_2a^{n-2}(bx)^2+\cdots+(bx)^n.

This is exact for every real $x$ . It is the printed $(a+b)^n$ identity with $bx$ replacing $b$ .

In the formula sheet.

The triangle also counts every possible success/failure pattern: row $n$ totals $2^n$ . For row $5$ , $1+5+10+10+5+1=32$ , the number of different strings of five binary outcomes. That doubling will explain a probability check later.

Worked example 3: a non-unit constant

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

Plan. Use row $4$ , decreasing the power of $2$ as the power of $x$ increases.

\begin{aligned} (2+x)^4&=2^4+4(2^3)x+6(2^2)x^2+4(2)x^3+x^4 &&\text{use all five coefficients}\\ &=16+32x+24x^2+8x^3+x^4 &&\text{simplify} \end{aligned}

Why is the constant $16$ ? At $x=0$ , all four brackets supply $2$ .

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

Check. At $x=1$ , the sum is $81=3^4$ .

Keep the exponent and constant, changing only the sign of $x$ .

Worked example 4: the signed variable term

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

Plan. Replace $x$ by $-x$ in the same finite expansion.

\begin{aligned} (2-x)^4&=2^4+4(2^3)(-x)+6(2^2)(-x)^2+4(2)(-x)^3+(-x)^4 &&\text{keep signs inside powers}\\ &=16-32x+24x^2-8x^3+x^4 &&\text{odd powers change sign} \end{aligned}

Why keep the even terms positive? An even number of negative factors has positive product.

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

Check. At $x=2$ , $16-64+96-64+16=0$ , matching $(2-2)^4$ .

The coefficients are the counts multiplied by the powers of the chosen terms.

Watch out

Wrong: $(2-x)^4=1-4x+6x^2-4x^3+x^4$ . This forgets the constant factors. Right: include $2^{4-r}$ in the term containing $x^r$ .

Change the magnitude of the variable term while retaining this structure.

Finding a particular term or coefficient

You can go straight to the required power. A contribution of $x^r$ comes from selecting $r$ variable terms and $n-r$ constants, so its coefficient is ${}^nC_r a^{n-r}b^r$ . This is extracted from the supplied expansion, rather than a separate formula to learn.

The term contains $x^r$ ; the coefficient is the number multiplying it. In ascending powers, the constant comes first and the $x^r$ term is in position $r+1$ .

Worked example 5: the opening puzzle

Question. Find the coefficient of $x^2$ in $(2+x)^6$ .

Plan. Select two variable factors and retain the four constant factors.

\begin{aligned} \text{coefficient of }x^2&={}^6C_2\,2^4\,1^2 &&\text{two selected positions}\\ &=15\times16=240 &&\text{include the constants} \end{aligned}

Why is $15$ insufficient? Each of the $15$ selections contributes $16x^2$ , not just $x^2$ .

Answer. The coefficient is $240$ , so the term is $240x^2$ .

Check. The six-bracket row starts $1,6,15$ , consistent with selecting two positions.

Now change only the requested power. Selecting an additional $x$ removes one factor of $2$ .

Worked example 6: the next power

Question. Find the coefficient of $x^3$ in $(2+x)^6$ .

Plan. Select three variable factors and three constant factors.

\begin{aligned} \text{coefficient of }x^3&={}^6C_3\,2^3 &&\text{three of each type}\\ &=20\times8=160 &&\text{evaluate} \end{aligned}

Why does the coefficient fall even though the count rises? Each individual contribution has halved, from $16$ to $8$ .

Answer. The coefficient is $160$ ; the term $160x^3$ is fourth.

Check. Counting ordered triples gives $(6\times5\times4)/3!=20$ , agreeing with row $6$ .

For a signed variable coefficient, include its sign inside the power. For example, the $x^3$ coefficient in $(3-2x)^6$ is ${}^6C_3 3^3(-2)^3=-4320$ . Its negativity comes from the odd power.

Exam tip

State the coefficient as a number. A displayed selection expression also shows which term you used, so a later arithmetic slip does not hide the method.

Change only the requested power again.

Why the same coefficients appear in probabilities

A string of successes and failures is another selection of positions. In four trials, exactly one success has four positions: SFFF, FSFF, FFSF and FFFS.

You already know

Independent trial probabilities multiply along one specified pattern. Mutually exclusive patterns have probabilities that add. With success probability $p$ , failure has probability $1-p$ .

If all trials have the same success probability, patterns with the same success count have equal probability. The binomial coefficient counts how many such patterns there are.

Binomial success count

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

\mathrm P(\text{exactly }r\text{ successes})={}^nC_r p^r(1-p)^{n-r},\qquad0\le r\le n.

Here $r$ is an integer. At $p=0$ , all failures have probability $1$ ; at $p=1$ , all successes have probability $1$ . Handle these endpoint cases directly, without substituting $0^0$ .

In the formula sheet.

The probability formula is the finite expansion with probability factors attached to each position selection.

Worked example 7: one success position

Question. Four independent trials each have success probability $0.2$ . Find the probability of exactly one success.

Plan. Calculate one pattern's probability, then count all possible success positions.

\begin{aligned} \mathrm P(\mathrm{SFFF})&=(0.2)(0.8)^3=0.1024 &&\text{multiply independent factors}\\ \mathrm P(\text{one success})&={}^4C_1(0.2)(0.8)^3 &&\text{four disjoint patterns}\\ &=0.4096 &&\text{add their probabilities} \end{aligned}

Why multiply by $4$ ? The success can occupy any of the four positions.

Answer. $0.4096$ .

Check. The four pattern probabilities are all $0.1024$ .

Change only the required count to two successes. The number of patterns now comes from pairs of positions.

Worked example 8: two success positions

Question. For the same four trials, find the probability of exactly two successes.

Plan. Multiply one two-success pattern's probability by ${}^4C_2$ .

\begin{aligned} \mathrm P(\mathrm{SSFF})&=(0.2)^2(0.8)^2=0.0256 &&\text{one specified pattern}\\ \mathrm P(\text{two successes})&=6(0.0256)=0.1536 &&\text{six position pairs} \end{aligned}

Why not $4$ patterns again? Two successes occupy a pair of positions, and ${}^4C_2=6$ .

Answer. $0.1536$ .

Check. For $0,1,2,3,4$ successes, the probabilities are $0.4096,0.4096,0.1536,0.0256,0.0016$ ; their sum is $1$ , matching $(0.8+0.2)^4$ .

If $p=1/2$ , every complete pattern has the same probability $1/2^n$ . Dividing a row total of $2^n$ by $2^n$ explains the total probability of $1$ directly.

Watch out

Wrong: use $p^r(1-p)^{n-r}$ for the complete success-count event. That is one pattern. Right: multiply by ${}^nC_r$ , provided the trials are independent and the success probability is constant.

Keep four trials and two successes, changing only the success probability.

Quick Recap

${}^nC_r$ counts unordered position selections; $0!=1$ keeps the endpoint counts equal to $1$ .
Pascal's triangle records the finite expansion coefficients, but the bracket's constant and variable factors still need their powers.
One selected coefficient can be found without a complete expansion.
In binomial probabilities, count success positions and then multiply by each pattern's probability.

The opening coefficient is $240$ : $15$ position selections, each contributing $2^4x^2=16x^2$ .

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Ms1, Ms2 - 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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