Intermediate

22 min read

Binomial Coefficients & Multinomial Theorem

Master Binomial Coefficients and Multinomial Theorem for JEE Main and Board Exams. Covers symmetry C(n,r)=C(n,n-r), Pascal identity C(n,r)+C(n,r+1)=C(n+1,r+1), sum identities C0+C1+...+Cn=2^n alternating sum=0 even odd coefficients 2^(n-1), weighted sum Σr·Cr=n·2^(n-1), sum of squares ΣCr²=C(2n,n), differentiation and integration methods, multinomial theorem coefficient n!/(r1!r2!...rk!), number of terms C(n+k-1,k-1), with fully verified JEE Board practice questions.

New

The Binomial Coefficients — the C(n,r) values that appear in the expansion of (a+b)ⁿ — are far more than just numbers in a formula. They satisfy a rich web of identities that connect them to combinatorics, to Pascal's Triangle, and to each other. These identities appear directly in JEE as "prove that" questions and MCQs involving sums of products of binomial coefficients. The Multinomial Theorem extends the Binomial Theorem to expansions with more than two terms — (a+b+c)ⁿ and beyond — and is tested through coefficient-finding problems. Both topics are compact but formulaically dense, rewarding students who memorise the key identities systematically.

1. Binomial Coefficients — Notation and Basic Properties

The binomial coefficient $(\binom{n}{r}) =^{n} C_{r} = \frac{n!}{r! (n - r)!}$ counts the number of ways to choose r items from n.

Fundamental Properties

Property	Formula	Example
Symmetry	C(n,r) = C(n, n−r)	C(10,3) = C(10,7) = 120
Boundary	C(n,0) = C(n,n) = 1	C(15,0) = 1
Pascal's Identity	C(n,r) + C(n,r+1) = C(n+1, r+1)	C(6,2)+C(6,3) = 15+20 = 35 = C(7,3) ✓
Recurrence	C(n,r) = (n/r)·C(n−1, r−1)	Builds triangle efficiently

2. Standard Sum Identities

All identities below follow from substituting special values of x in the expansion of (1+x)ⁿ.

Identity	Formula	Obtained by
Sum of all coeff.	C₀ + C₁ + C₂ + ··· + Cₙ = 2ⁿ	Put x = 1 in (1+x)ⁿ
Alternating sum	C₀ − C₁ + C₂ − ··· = 0	Put x = −1 in (1+x)ⁿ
Even-position coeff.	C₀ + C₂ + C₄ + ··· = 2^(n−1)	Average of x=1 and x=−1 results
Odd-position coeff.	C₁ + C₃ + C₅ + ··· = 2^(n−1)	Same as above
Weighted sum	1·C₁ + 2·C₂ + 3·C₃ + ··· + n·Cₙ = n·2^(n−1)	Differentiate (1+x)ⁿ and put x=1
Sum of squares	C₀² + C₁² + C₂² + ··· + Cₙ² = C(2n, n)	Product of (1+x)ⁿ and (x+1)ⁿ
Cross product	C₀Cᵣ + C₁Cᵣ₊₁ + ··· + Cₙ₋ᵣCₙ = C(2n, n−r)	Generalisation of sum of squares

Verifications

n = 10: C₀²+C₁²+···+C₁₀² = C(20,10) = 184,756 ✓

n = 10: 1·C₁+2·C₂+···+10·C₁₀ = 10·2⁹ = 10·512 = 5120 ✓

3. Deriving Identities — Differentiation and Integration Tricks

Differentiation Method

Start with (1+x)ⁿ = C₀ + C₁x + C₂x² + ··· + Cₙxⁿ

Differentiate: n(1+x)^(n−1) = C₁ + 2C₂x + 3C₃x² + ··· + nCₙx^(n−1)

Put x = 1: n·2^(n−1) = C₁ + 2C₂ + 3C₃ + ··· + nCₙ

Put x = −1: 0 = C₁ − 2C₂ + 3C₃ − ··· (for n ≥ 2)

Integration Method

Integrate (1+x)ⁿ = C₀ + C₁x + ··· + Cₙxⁿ from 0 to 1:

2^(n+1)/(n+1) − 1/(n+1) = C₀ + C₁/2 + C₂/3 + ··· + Cₙ/(n+1)

So: C₀ + C₁/2 + C₂/3 + ··· + Cₙ/(n+1) = (2^(n+1) − 1)/(n+1)

4. Multinomial Theorem

The expansion of (a₁ + a₂ + ··· + aₖ)ⁿ is:

$(a_{1} + a_{2} + \dots + a_{k})^{n} = \sum_{r_{1} + r_{2} + \dots + r_{k} = n} \frac{n!}{r_{1}! r_{2}! \dots r_{k}!} a_{1}^{r_{1}} a_{2}^{r_{2}} \dots a_{k}^{r_{k}}$

Number of Terms

$Number of terms in (a_{1} + a_{2} + \dots + a_{k})^{n} = (\binom{n + k - 1}{k - 1})$

Expansion	Number of terms
(a+b)ⁿ — 2 terms	C(n+1,1) = n + 1
(a+b+c)ⁿ — 3 terms	C(n+2,2) = (n+1)(n+2)/2
(a+b+c+d)ⁿ — 4 terms	C(n+3,3)

Finding a Specific Coefficient in a Multinomial

The coefficient of a₁^r₁ · a₂^r₂ · ··· · aₖ^rₖ (where r₁+r₂+···+rₖ = n) is:

$Coefficient = \frac{n!}{r_{1}! r_{2}! \dots r_{k}!}$

Worked Examples

Example 1: Find the coefficient of a²bc in (a + b + c)⁴.

r₁ = 2, r₂ = 1, r₃ = 1; r₁+r₂+r₃ = 4 ✓

Coefficient = 4!/(2!·1!·1!) = 24/2 = 12

Example 2: Find the coefficient of a²b²c in (2a + 3b − c)⁵.

r₁=2, r₂=2, r₃=1; coefficient = 5!/(2!·2!·1!) × (2)² × (3)² × (−1)¹

= 30 × 4 × 9 × (−1) = −1080

5. Ratio of Consecutive Binomial Coefficients

$\frac{C (n, r)}{C (n, r - 1)} = \frac{n - r + 1}{r}$

This ratio is used to find the greatest term, and to solve problems where consecutive coefficients are in given ratios.

Example: If C(n,3)/C(n,2) = 5, then (n−2)/3 = 5 → n = 17.

Binomial Coefficient Identities — Summary for Exams

What to find	Method	Result
Sum of all Cᵣ	x=1 in (1+x)ⁿ	2ⁿ
Alternating sum	x=−1 in (1+x)ⁿ	0
Σr·Cᵣ	Differentiate, x=1	n·2^(n−1)
Σ Cᵣ²	Coefficient of xⁿ in (1+x)^(2n)	C(2n,n)
Σ Cᵣ/(r+1)	Integrate, 0 to 1	(2^(n+1) − 1)/(n+1)
Multinomial coeff.	n!/(r₁!r₂!···rₖ!)	Include base values raised to powers

JEE/Board Power Tips

C₀+C₁+···+Cₙ = 2ⁿ is for (1+x)ⁿ specifically. For (a+b)ⁿ, the sum of all binomial coefficients C(n,r) is still 2ⁿ, but the sum of all terms (including the a and b factors) is (a+b)ⁿ evaluated at a=b=1, which is 2ⁿ only if a=b=1.
Sum of squares Σ Cᵣ² = C(2n,n). This comes from the coefficient of xⁿ in (1+x)ⁿ × (1+x)ⁿ = (1+x)^(2n). It is a frequently tested JEE result.
For multinomial coefficients: always verify r₁+r₂+···+rₖ = n before computing. If the powers don't sum to n, that term doesn't exist in the expansion.
Pascal's Identity C(n,r) + C(n,r+1) = C(n+1,r+1) is the foundation of Pascal's Triangle. Each entry is the sum of the two entries directly above it — a direct consequence of this identity.

Practice Questions

Q1 (Board): Prove that $C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}$ .

Explanation:

Consider the standard binomial expansion of $(1 + x)^{n}$ :
$(1 + x)^{n} = C_{0} + C_{1} x + C_{2} x^{2} + \dots + C_{n} x^{n}$

To find the sum of all the binomial coefficients, simply substitute $x = 1$ into the identity:
$(1 + 1)^{n} = C_{0} + C_{1} (1) + C_{2} (1)^{2} + \dots + C_{n} (1)^{n}$

Therefore, $C_{0} + C_{1} + C_{2} + \dots + C_{n} = 2^{n}$ . $◼$

Q2 (JEE Main level): Find the value of $1 \cdot C_{1} + 2 \cdot C_{2} + 3 \cdot C_{3} + \dots + n \cdot C_{n}$ for $n = 5$ .

Explanation:

Using the standard binomial derivative identity:
$\sum_{r = 1}^{n} r \cdot C_{r} = n \cdot 2^{n - 1}$

Substitute $n = 5$ into the formula:
$1 \cdot C_{1} + 2 \cdot C_{2} + 3 \cdot C_{3} + 4 \cdot C_{4} + 5 \cdot C_{5} = 5 \cdot 2^{5 - 1} = 5 \cdot 2^{4}$
$= 5 \cdot 16 = 80$

Verification by direct expansion:
$C_{1} = 5, C_{2} = 10, C_{3} = 10, C_{4} = 5, C_{5} = 1$
$1 (5) + 2 (10) + 3 (10) + 4 (5) + 5 (1) = 5 + 20 + 30 + 20 + 5 = 80$ ✓

Q3 (Board / JEE): Find the coefficient of $x^{2} y^{3} z$ in the expansion of $(2 x - y + 3 z)^{6}$ .

Explanation:

Using the Multinomial Theorem, the general term for $(a + b + c)^{n}$ is:
$\frac{n!}{r_{1}! \cdot r_{2}! \cdot r_{3}!} a^{r_{1}} b^{r_{2}} c^{r_{3}}$

Here, $n = 6$ , $a = 2 x$ , $b = - y$ , and $c = 3 z$ .
We want the term containing $x^{2} y^{3} z^{1}$ , so we set $r_{1} = 2$ , $r_{2} = 3$ , and $r_{3} = 1$ .
(Check: $r_{1} + r_{2} + r_{3} = 2 + 3 + 1 = 6 = n$ ✓)

Substitute the values:
$Term = \frac{6!}{2! \cdot 3! \cdot 1!} (2 x)^{2} (- y)^{3} (3 z)^{1}$
$Term = (\frac{720}{2 \cdot 6 \cdot 1}) \cdot (4 x^{2}) \cdot (- 1 y^{3}) \cdot (3 z)$
$Term = 60 \cdot 4 \cdot (- 1) \cdot 3 \cdot x^{2} y^{3} z$
$Term = 60 \cdot (- 12) \cdot x^{2} y^{3} z = - 720 x^{2} y^{3} z$

The coefficient is $- 720$ .

Q4 (MCQ — 1 mark): The value of $C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + C_{3}^{2}$ where $C_{r} = (\binom{3}{r})$ equals:

A) 8
B) 20
C) 16
D) 12

Answer: B) 20.

Explanation: The sum of the squares of binomial coefficients is given by the identity:
$\sum_{r = 0}^{n} C_{r}^{2} = (\binom{2 n}{n})$

Here, $n = 3$ .
Sum = $(\binom{2 (3)}{3}) = (\binom{6}{3}) = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20$ .

Verification: $1^{2} + 3^{2} + 3^{2} + 1^{2} = 1 + 9 + 9 + 1 = 20$ ✓

Q5 (Board): Find the total number of distinct terms in the expansion of $(a + b + c + d)^{5}$ .

Explanation:

The number of distinct terms in the expansion of a multinomial $(x_{1} + x_{2} + \dots + x_{k})^{n}$ is calculated using the "stars and bars" combinatorics formula:
$Number of terms = (\binom{n + k - 1}{k - 1})$

Here, the exponent $n = 5$ and the number of variables inside the bracket $k = 4$ ( $a, b, c, d$ ).
$Number of terms = (\binom{5 + 4 - 1}{4 - 1}) = (\binom{8}{3})$

Evaluate $(\binom{8}{3})$ :
$(\binom{8}{3}) = \frac{8!}{3! (8 - 3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$

Related Study Material

More Mathematics Notes More Algebra Notes All Binomial Theorem Topics

Practice Questions

Mathematics Practice Binomial Theorem Questions

Topic Information

DifficultyIntermediate

Est. Read Time22 minutes

CategoryBinomial Theorem

Quick Actions

Track Your Learning

Study Tips

Take notes of key points while reading

Practice related questions after studying

Review topics marked as "Learning" regularly

Test your knowledge with practice questions

Edvaya Target

Detailed Notes & Solved Examples

Master complex concepts with our rich, topic-wise study notes. Access hundreds of step-by-step solved examples and interactive flashcards right from your phone.

Edvaya Target

Edvaya Aspire

Take a Free Mock Test

Target Subjects

Aspire Subjects

Subject Mastery

Binomial Coefficients & Multinomial Theorem

1. Binomial Coefficients — Notation and Basic Properties

Fundamental Properties

2. Standard Sum Identities

Verifications

3. Deriving Identities — Differentiation and Integration Tricks

Differentiation Method

Integration Method

4. Multinomial Theorem

Number of Terms

Finding a Specific Coefficient in a Multinomial

Worked Examples

5. Ratio of Consecutive Binomial Coefficients

Binomial Coefficient Identities — Summary for Exams

Practice Questions

Related Study Material

Practice Questions

Topic Information

Quick Actions

Track Your Learning

Study Tips

Edvaya Target

We Value Your Privacy