Binomial methods

This chapter provides a rigorous treatment of binomial methods, a fundamental topic in combinatorics and discrete mathematics. Mastery of binomial expansions, coefficients, and identities is crucial for solving a wide array of problems frequently encountered in CMI examinations, particularly those involving enumeration and probability.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Binomial theorem | | 2 | Coefficients | | 3 | Special expansions | | 4 | Binomial identities |

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We begin with Binomial theorem.

Part 1: Binomial theorem

Binomial Theorem

Overview

The binomial theorem is one of the most useful expansion tools in algebra and combinatorics. It tells us how to expand powers of a binomial such as $(a+b)^n$ in a precise coefficient pattern. In CMI-style questions, the theorem is rarely tested as a plain expansion exercise; instead, it appears in coefficient extraction, term identification, divisibility, parity, approximation, and combinatorial interpretation. ---

Learning Objectives

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Core Statement

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General Term

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Expansions with Signs

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Important Properties of Binomial Coefficients

These identities are fundamental in both algebra and counting. ---

Middle Term(s)

Example: In $(a+b)^8$, the middle term is the $5$th term. In $(a+b)^7$, the middle terms are the $4$th and $5$th terms. ---

How Powers Change Across Terms

This is essential when finding a particular term or matching a required power. ---

Minimal Worked Examples

Example 1 Find the general term in $(2x-3)^7$. Here $a=2x$ and $b=-3$. So $\qquad T_{r+1}=\binom{7}{r}(2x)^{7-r}(-3)^r$ for $r=0,1,2,\dots,7$. --- Example 2 Find the coefficient of $x^3$ in $(1+x)^5$. The general term is $\qquad T_{r+1}=\binom{5}{r}x^r$ To get $x^3$, take $r=3$. So the coefficient is $\qquad \binom{5}{3}=10$ Hence the coefficient is $\boxed{10}$. ---

Coefficient Interpretation

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Common Patterns in Questions

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The coefficient of $x^2$ in $(1+x)^4$ is" options=["$4$","$6$","$8$","$12$"] answer="B" hint="Use the term $\binom{4}{2}x^2$." solution="In the expansion of $(1+x)^4$, the general term is $\qquad \binom{4}{r}x^r$. For the coefficient of $x^2$, take $r=2$. So the coefficient is $\qquad \binom{4}{2}=6$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the coefficient of $x^3$ in $(2+x)^5$." answer="40" hint="Use the general term." solution="The general term is $\qquad T_{r+1}=\binom{5}{r}2^{\,5-r}x^r$. To get $x^3$, take $r=3$. So the coefficient is $\qquad \binom{5}{3}2^2=10\cdot 4=40$. Therefore the answer is $\boxed{40}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\binom{n}{r}=\binom{n}{n-r}$","The sum of all coefficients in $(1+x)^n$ is $2^n$","The exponent of $a$ increases as the exponent of $b$ increases in $(a+b)^n$","In $(a-b)^n$, signs may alternate"] answer="A,B,D" hint="Recall the standard expansion pattern." solution="1. True, by symmetry of binomial coefficients.

True, putting $x=1$ in $(1+x)^n$ gives $2^n$.

False. In $(a+b)^n$, as the exponent of $b$ increases, the exponent of $a$ decreases.

True. In $(a-b)^n$, the factor $(-b)^r$ introduces alternating signs.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find the middle term of the expansion of $(x+2)^8$." answer="$1120x^4$" hint="For even $n$, there is one middle term." solution="Since $n=8$ is even, there is one middle term, namely the $\left(\dfrac{8}{2}+1\right)=5$th term. The general term is $\qquad T_{r+1}=\binom{8}{r}x^{8-r}2^r$ For the $5$th term, we take $r=4$. So $\qquad T_5=\binom{8}{4}x^4 2^4$ Now $\qquad \binom{8}{4}=70,\qquad 2^4=16$ Hence $\qquad T_5=70\cdot 16\,x^4=1120x^4$ Therefore the middle term is $\boxed{1120x^4}$." ::: ---

Summary

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Part 2: Coefficients

Coefficients

Overview

Coefficient problems ask for the numerical multiplier of a particular term in an algebraic expansion or expression. In binomial-method questions, coefficients are often extracted from expansions such as $(1+x)^n$, $(a+bx)^n$, or products of several series. The difficulty is usually not the expansion itself, but matching the required power carefully and distinguishing between a coefficient and a full term. ---

Learning Objectives

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Core Idea

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Coefficient from Binomial Expansion

So the coefficient of $x^r$ is $\qquad \binom{n}{r}a^{\,n-r}b^r$ ::: ---

Matching the Required Power

This is the most reliable method. ---

Coefficient in Expressions Like $(x+a)^n$

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Constant Term Problems

Example idea: In $(x+\dfrac{1}{x})^6$, the general term is $\qquad T_{r+1}=\binom{6}{r}x^{6-r}\left(\dfrac{1}{x}\right)^r=\binom{6}{r}x^{6-2r}$ The constant term comes when $\qquad 6-2r=0 \implies r=3$ So the constant term is $\qquad \binom{6}{3}=20$ ---

Coefficients in Products

This second method is often the combinatorial route. ---

Minimal Worked Examples

Example 1 Find the coefficient of $x^2$ in $(3+x)^4$. General term: $\qquad T_{r+1}=\binom{4}{r}3^{4-r}x^r$ For $x^2$, take $r=2$. Coefficient: $\qquad \binom{4}{2}3^2=6\cdot 9=54$ So the coefficient is $\boxed{54}$. --- Example 2 Find the constant term in $(x+\dfrac{1}{x})^4$. General term: $\qquad T_{r+1}=\binom{4}{r}x^{4-r}\left(\dfrac{1}{x}\right)^r=\binom{4}{r}x^{4-2r}$ For the constant term, set $\qquad 4-2r=0 \implies r=2$ So the constant term is $\qquad \binom{4}{2}=6$ Hence the answer is $\boxed{6}$. ---

Important Distinctions

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Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The coefficient of $x^3$ in $(1+x)^5$ is" options=["$5$","$10$","$15$","$20$"] answer="B" hint="Use $\binom{5}{3}$." solution="In $(1+x)^5$, the coefficient of $x^r$ is $\binom{5}{r}$. For $x^3$, the coefficient is $\qquad \binom{5}{3}=10$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the constant term in $\left(x+\dfrac{1}{x}\right)^6$." answer="20" hint="Set the exponent of $x$ equal to zero." solution="The general term is $\qquad T_{r+1}=\binom{6}{r}x^{6-r}\left(\dfrac{1}{x}\right)^r=\binom{6}{r}x^{6-2r}$ For the constant term, set $\qquad 6-2r=0$ So $\qquad r=3$ Hence the constant term is $\qquad \binom{6}{3}=20$ Therefore the answer is $\boxed{20}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The constant term is the coefficient of $x^0$","In $(a+bx)^n$, the coefficient of $x^r$ is $\binom{n}{r}a^{n-r}b^r$","The term and the coefficient are always the same thing","To extract a coefficient, matching the power of $x$ is usually the key step"] answer="A,B,D" hint="Recall the general term and the meaning of constant term." solution="1. True. By definition, the constant term is the coefficient of $x^0$.

True. This follows directly from the binomial general term.

False. A term includes the power of $x$, while the coefficient is only its multiplier.

True. Power matching is the standard method.

Hence the correct answer is $\boxed{A,B,D}$." ::: :::question type="SUB" question="Find the coefficient of $x^2$ in $(2+3x)^4$." answer="$216$" hint="Write the general term and take $r=2$." solution="The general term is $\qquad T_{r+1}=\binom{4}{r}2^{4-r}(3x)^r$ To get $x^2$, take $r=2$. So the coefficient is $\qquad \binom{4}{2}2^2 3^2$ $\qquad =6\cdot 4\cdot 9=216$ Therefore the coefficient is $\boxed{216}$." ::: ---

Summary

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Part 3: Special expansions

Special Expansions

Overview

Special expansions are algebraic identities that appear repeatedly in counting, simplification, coefficient extraction, and proof-based questions. In CMI-style problems, these formulas are not used mechanically; the key is to recognise the right expansion or factorisation at the right moment. ---

Learning Objectives

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Core Formulas

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High-Value Derived Identities

These are often faster than direct expansion. ---

Pattern Recognition

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Minimal Worked Examples

Example 1 Simplify $\qquad (x+2)^2 - (x-2)^2$ Using the square expansions, $\qquad (x+2)^2 = x^2+4x+4$ $\qquad (x-2)^2 = x^2-4x+4$ Subtracting, $\qquad (x+2)^2-(x-2)^2 = 8x$ --- Example 2 Simplify $\qquad a^3+b^3-(a+b)^3$ Using $\qquad (a+b)^3 = a^3+3a^2b+3ab^2+b^3$ we get $\qquad a^3+b^3-(a+b)^3 = -3a^2b-3ab^2 = -3ab(a+b)$ ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Which of the following is equal to $(a-b)^2$?" options=["$a^2-b^2$","$a^2-2ab+b^2$","$a^2+2ab+b^2$","$a^2-ab+b^2$"] answer="B" hint="Recall the square expansion exactly." solution="The standard identity is $\qquad (a-b)^2 = a^2-2ab+b^2$ Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the value of $(51)^2-(49)^2$." answer="200" hint="Use the identity $a^2-b^2=(a-b)(a+b)$." solution="Using $\qquad a^2-b^2=(a-b)(a+b)$, we get $\qquad 51^2-49^2=(51-49)(51+49)=2\cdot 100=200$ Hence the answer is $\boxed{200}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$(a+b)^2=a^2+2ab+b^2$","$a^3-b^3=(a-b)(a^2+ab+b^2)$","$a^3+b^3=(a+b)(a^2-ab+b^2)$","$(a+b)(a-b)=a^2+b^2$"] answer="A,B,C" hint="Check each identity carefully." solution="1. True.

True.

True.

False, because

$\qquad (a+b)(a-b)=a^2-b^2$ Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that $(a+b)^2+(a-b)^2=2(a^2+b^2)$." answer="$2(a^2+b^2)$" hint="Expand both squares and combine like terms." solution="Expand both terms: $\qquad (a+b)^2 = a^2+2ab+b^2$ $\qquad (a-b)^2 = a^2-2ab+b^2$ Adding, $\qquad (a+b)^2+(a-b)^2 = (a^2+2ab+b^2)+(a^2-2ab+b^2)$ $\qquad = 2a^2+2b^2 = 2(a^2+b^2)$ Hence the identity is proved, and the result is $\boxed{2(a^2+b^2)}$." ::: ---

Summary

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Part 4: Binomial identities

Binomial Identities

Overview

Binomial identities are algebraic identities involving binomial coefficients. They arise from expanding $(1+x)^n$, comparing coefficients, differentiating the binomial expansion, and interpreting coefficients combinatorially. In CMI-style problems, the real skill is not memorizing many formulas mechanically, but knowing where each identity comes from and when to use it. ---

Learning Objectives

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Core Definition

Almost all standard binomial identities come from substituting convenient values of $x$ or differentiating this formula. ---

Essential Identities

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Where These Come From

1. Set $x=1$

From $\qquad (1+x)^n = \sum_{r=0}^n \binom{n}{r}x^r$ put $x=1$: $\qquad 2^n = \sum_{r=0}^n \binom{n}{r}$ ---

2. Set $x=-1$

Put $x=-1$: $\qquad (1-1)^n = \sum_{r=0}^n \binom{n}{r}(-1)^r$ So for $n\ge 1$, $\qquad 0 = \sum_{r=0}^n (-1)^r \binom{n}{r}$ ---

3. Differentiate

Differentiate $\qquad (1+x)^n$ to get $\qquad n(1+x)^{n-1} = \sum_{r=1}^{n} r\binom{n}{r}x^{r-1}$ Now set $x=1$: $\qquad n2^{n-1} = \sum_{r=1}^{n} r\binom{n}{r}$ This is one of the most frequently used identities. ---

Combinatorial Meaning

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More Useful Identities

These appear in stronger counting and algebra problems. ---

Minimal Worked Examples

Example 1 Evaluate $\qquad \sum_{r=0}^{6}\binom{6}{r}$ Using the identity, $\qquad \sum_{r=0}^{n}\binom{n}{r}=2^n$ we get $\qquad \sum_{r=0}^{6}\binom{6}{r} = 2^6 = 64$ So the answer is $\boxed{64}$. --- Example 2 Evaluate $\qquad \sum_{r=0}^{5} r\binom{5}{r}$ Using $\qquad \sum_{r=0}^{n} r\binom{n}{r}=n2^{n-1}$, we get $\qquad \sum_{r=0}^{5} r\binom{5}{r}=5\cdot 2^4=80$ So the answer is $\boxed{80}$. ---

Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="Evaluate $\sum_{r=0}^{8}\binom{8}{r}$." options=["$64$","$128$","$256$","$512$"] answer="C" hint="Use the sum of all binomial coefficients." solution="Using $\qquad \sum_{r=0}^{n}\binom{n}{r}=2^n$, we get $\qquad \sum_{r=0}^{8}\binom{8}{r}=2^8=256$. Hence the correct option is $\boxed{C}$." ::: :::question type="NAT" question="Find $\sum_{r=0}^{6} r\binom{6}{r}$." answer="192" hint="Use the differentiated form of the binomial theorem." solution="Using the identity $\qquad \sum_{r=0}^{n} r\binom{n}{r}=n2^{n-1}$, with $n=6$, we get $\qquad \sum_{r=0}^{6} r\binom{6}{r}=6\cdot 2^5=6\cdot 32=192$. Hence the answer is $\boxed{192}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["$\binom{n}{r}=\binom{n}{n-r}$","$\sum_{r=0}^{n}\binom{n}{r}=2^n$","$\sum_{r=0}^{n}(-1)^r\binom{n}{r}=0$ for $n\ge 1$","$r\binom{n}{r}=\binom{n-1}{r-1}$"] answer="A,B,C" hint="One identity is missing a factor." solution="1. True. This is the symmetry identity.

True. It follows from putting $x=1$ in $(1+x)^n$.

True. It follows from putting $x=-1$ in $(1+x)^n$ for $n\ge 1$.

False. The correct identity is

$\qquad r\binom{n}{r}=n\binom{n-1}{r-1}$. Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Prove that $\sum_{r=0}^{n}\binom{n}{r}=2^n$." answer="By setting $x=1$ in the binomial theorem." hint="Start from $(1+x)^n$." solution="From the binomial theorem, $\qquad (1+x)^n=\sum_{r=0}^{n}\binom{n}{r}x^r$ Now put $x=1$. Then $\qquad (1+1)^n=\sum_{r=0}^{n}\binom{n}{r}$ So $\qquad 2^n=\sum_{r=0}^{n}\binom{n}{r}$ Hence the identity is proved." ::: ---

Summary

Chapter Summary

Chapter Review Questions

:::question type="MCQ" question="What is the coefficient of $x^7$ in the expansion of $(2x^2 - \frac{1}{x})^8$?" options=["-1792","1792","-896","896"] answer="-1792" hint="Identify the general term $\binom{n}{k} (ax^p)^{n-k} (bx^q)^k$ and equate the exponent of $x$ to 7." solution="The general term is $T_{k+1} = \binom{8}{k} (2x^2)^{8-k} \left(-\frac{1}{x}\right)^k = \binom{8}{k} 2^{8-k} x^{2(8-k)} (-1)^k x^{-k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16-2k-k} = \binom{8}{k} 2^{8-k} (-1)^k x^{16-3k}$.
For the coefficient of $x^7$, we set $16-3k = 7$, which implies $3k=9$, so $k=3$.
Substituting $k=3$ into the coefficient part: $\binom{8}{3} 2^{8-3} (-1)^3 = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} \times 2^5 \times (-1) = 56 \times 32 \times (-1) = -1792$."
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:::question type="NAT" question="Evaluate the sum:

" answer="5120" hint="Recall the identity for $\sum_{k=0}^n k \binom{n}{k}$ or differentiate the binomial expansion of $(1+x)^n$ with respect to $x$ and then substitute $x=1$." solution="We know the identity $\sum_{k=0}^n k \binom{n}{k} = n 2^{n-1}$.
For $n=10$, the sum is $10 \times 2^{10-1} = 10 \times 2^9 = 10 \times 512 = 5120$.
Alternatively, consider the expansion $(1+x)^{10} = \sum_{k=0}^{10} \binom{10}{k} x^k$.
Differentiating with respect to $x$: $10(1+x)^9 = \sum_{k=0}^{10} k \binom{10}{k} x^{k-1}$.
Setting $x=1$: $10(1+1)^9 = \sum_{k=0}^{10} k \binom{10}{k} (1)^{k-1}$.
$10 \times 2^9 = \sum_{k=0}^{10} k \binom{10}{k}$.
$10 \times 512 = 5120$."
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:::question type="MCQ" question="Which of the following represents the first three terms in the expansion of $(1-2x)^{1/2}$ for $|x| < 1/2$?" options=["$1 - x - \frac{1}{2} x^2$","$1 - x + \frac{1}{2} x^2$","$1 - 2x + 2x^2$","$1 - x - x^2$"] answer="$1 - x - \frac{1}{2} x^2$" hint="Use the generalized binomial theorem: $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!} y^2 + \dots$" solution="Using the generalized binomial theorem $(1+y)^n = 1 + ny + \frac{n(n-1)}{2!} y^2 + \dots$
Here, $y = -2x$ and $n = 1/2$.
The first three terms are:
$1 + \left(\frac{1}{2}\right)(-2x) + \frac{\left(\frac{1}{2}\right)\left(\frac{1}{2}-1\right)}{2!} (-2x)^2$
$= 1 - x + \frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2} (4x^2)$
$= 1 - x + \frac{-\frac{1}{4}}{2} (4x^2)$
$= 1 - x - \frac{1}{8} (4x^2)$
$= 1 - x - \frac{1}{2} x^2$."
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:::question type="NAT" question="What is the value of $\binom{12}{5} + \binom{12}{6}$?" answer="1716" hint="Apply Pascal's Identity: $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$." solution="Using Pascal's Identity, $\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}$.
Here, $n=12$ and $k=5$.
So, $\binom{12}{5} + \binom{12}{6} = \binom{12+1}{5+1} = \binom{13}{6}$.
$\binom{13}{6} = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{6 \times 5 \times 4 \times 3 \times 2 \times 1} = 13 \times 11 \times 3 \times 4 / (5 \times 1) = 13 \times 11 \times 3 \times 2 \times 4 / (6 \times 1) = 13 \times 11 \times 2 \times (10 \times 9 \times 8) / (5 \times 4 \times 3 \times 2 \times 1) = 13 \times 11 \times 2 \times 3 \times 2 \times 1 = 13 \times 11 \times 12 = 1716$.
Calculation:
$\binom{13}{6} = \frac{13 \times 12 \times 11 \times 10 \times 9 \times 8}{720}$
$= 13 \times (12/6/2) \times 11 \times (10/5) \times (9/3) \times (8/4)$
$= 13 \times 1 \times 11 \times 2 \times 3 \times 2$
$= 13 \times 11 \times 12 = 143 \times 12 = 1716$."
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What's Next?