Properties of Binomial Coefficients
Complete guide to binomial coefficient properties, identities, and applications in JEE problems
Properties of Binomial Coefficients
Introduction
Binomial coefficients are the numerical coefficients in the expansion of $(a + b)^n$, denoted by $\binom{n}{r}$ or ${}^nC_r$. They have remarkable properties that simplify many algebraic calculations.
Interactive Demo: Pascal’s Triangle
Build Pascal's Triangle
Each number is the sum of the two numbers above it. This visualizes Pascal's Identity: C(n,r) = C(n-1,r-1) + C(n-1,r)
Definition
$$\boxed{\binom{n}{r} = {}^nC_r = \frac{n!}{r!(n-r)!}}$$
where $0 \leq r \leq n$ and $n, r$ are non-negative integers.
Special Cases:
- $\binom{n}{0} = 1$
- $\binom{n}{n} = 1$
- $\binom{n}{1} = n$
- $\binom{n}{n-1} = n$
Fundamental Properties
Interactive Demo: Visualize Coefficient Properties
Explore properties like symmetry through Pascal’s Triangle.
Property 1: Symmetry Property
$$\boxed{\binom{n}{r} = \binom{n}{n-r}}$$
Proof:
$$\binom{n}{r} = \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)![n-(n-r)]!} = \binom{n}{n-r}$$Example: $\binom{10}{3} = \binom{10}{7} = 120$
Property 2: Pascal’s Identity
$$\boxed{\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}}$$
Proof:
$$\begin{align} \binom{n-1}{r-1} + \binom{n-1}{r} &= \frac{(n-1)!}{(r-1)!(n-r)!} + \frac{(n-1)!}{r!(n-r-1)!} \\ &= \frac{(n-1)!}{(r-1)!(n-r-1)!}\left[\frac{1}{n-r} + \frac{1}{r}\right] \\ &= \frac{(n-1)!}{(r-1)!(n-r-1)!} \cdot \frac{n}{r(n-r)} \\ &= \frac{n!}{r!(n-r)!} = \binom{n}{r} \end{align}$$This property generates Pascal’s Triangle.
Property 3: Sum of All Binomial Coefficients
$$\boxed{\binom{n}{0} + \binom{n}{1} + \binom{n}{2} + \cdots + \binom{n}{n} = 2^n}$$
Proof: Put $a = b = 1$ in $(a+b)^n$:
$$(1+1)^n = \sum_{r=0}^{n}\binom{n}{r} = 2^n$$Example: $\binom{5}{0} + \binom{5}{1} + \binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5} = 1+5+10+10+5+1 = 32 = 2^5$
Property 4: Sum of Coefficients at Odd and Even Places
$$\boxed{\binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots = 2^{n-1}}$$
Proof: Put $a = 1, b = -1$ in $(a+b)^n$:
$$(1-1)^n = \binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots = 0$$Let $E = \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + \cdots$ (even positions)
Let $O = \binom{n}{1} + \binom{n}{3} + \binom{n}{5} + \cdots$ (odd positions)
Then: $E - O = 0 \Rightarrow E = O$
And: $E + O = 2^n \Rightarrow 2E = 2^n \Rightarrow E = O = 2^{n-1}$
Property 5: Sum with Variable Powers
$$\boxed{\binom{n}{0} - \binom{n}{1} + \binom{n}{2} - \binom{n}{3} + \cdots + (-1)^n\binom{n}{n} = 0}$$
Proof: Put $x = -1$ in $(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}x^r$
Property 6: Ratio of Consecutive Coefficients
$$\boxed{\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}}$$
Proof:
$$\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n!/(r!(n-r)!)}{n!/((r-1)!(n-r+1)!)} = \frac{(r-1)!(n-r+1)!}{r!(n-r)!} = \frac{n-r+1}{r}$$Advanced Properties
Property 7: Sum of Products
$$\boxed{\binom{n}{0}^2 + \binom{n}{1}^2 + \binom{n}{2}^2 + \cdots + \binom{n}{n}^2 = \binom{2n}{n}}$$
Proof: Consider $(1+x)^n \cdot (1+x)^n = (1+x)^{2n}$
Coefficient of $x^n$ on LHS = $\sum_{r=0}^{n}\binom{n}{r}\binom{n}{n-r} = \sum_{r=0}^{n}\binom{n}{r}^2$
Coefficient of $x^n$ on RHS = $\binom{2n}{n}$
Property 8: Sum with Index Multiplication
$$\boxed{\sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1}}$$
Proof:
$$\sum_{r=0}^{n} r \cdot \binom{n}{r} x^r = x\frac{d}{dx}[(1+x)^n] = x \cdot n(1+x)^{n-1}$$Put $x = 1$: $\sum_{r=0}^{n} r \cdot \binom{n}{r} = n \cdot 2^{n-1}$
Property 9: Vandermonde’s Identity
$$\boxed{\sum_{r=0}^{k} \binom{m}{r}\binom{n}{k-r} = \binom{m+n}{k}}$$
This is the coefficient of $x^k$ in $(1+x)^m(1+x)^n = (1+x)^{m+n}$
Memory Tricks
🎯 Symmetry Memory
“Mirror Property”: $\binom{n}{r} = \binom{n}{n-r}$
- Choosing $r$ items = Leaving $(n-r)$ items
- Both are the same!
🎯 Pascal’s Triangle Builder
“Add the neighbors above”:
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
Each number = sum of two numbers directly above it.
🎯 Sum Formula Memory
“Power of 2 Rule”:
- All coefficients sum: $2^n$
- Even position sum = Odd position sum = $2^{n-1}$
- Think: “Half the total”
🎯 Ratio Trick
“Top-Bottom Rule”: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$
Mnemonic: “New Minus R plus 1, over R”
Common Mistakes to Avoid
❌ Mistake 1: Wrong Symmetry Application
Wrong: $\binom{10}{3} = \binom{3}{10}$ ✗
Correct: $\binom{10}{3} = \binom{10}{10-3} = \binom{10}{7}$ ✓
Remember: Both $n$ values must be same in symmetry!
❌ Mistake 2: Pascal’s Identity Confusion
Wrong: $\binom{n}{r} = \binom{n}{r-1} + \binom{n}{r+1}$ ✗
Correct: $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ ✓
Remember: Go to previous row (n-1), not same row!
❌ Mistake 3: Sum Calculation Error
Wrong: Sum of odd-positioned coefficients = $2^n$ ✗
Correct: Sum of odd-positioned OR even-positioned = $2^{n-1}$ ✓
❌ Mistake 4: Ratio Formula Misuse
Wrong: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{r}{n-r+1}$ ✗
Correct: $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$ ✓
Solved Examples
Example 1: Using Symmetry (JEE Main)
Evaluate: $\binom{20}{17} + \binom{20}{16}$
Solution: Using symmetry: $\binom{20}{17} = \binom{20}{3}$ and $\binom{20}{16} = \binom{20}{4}$
$$\begin{align} \binom{20}{3} + \binom{20}{4} &= \frac{20 \times 19 \times 18}{3 \times 2 \times 1} + \frac{20 \times 19 \times 18 \times 17}{4 \times 3 \times 2 \times 1} \\ &= 1140 + 4845 \\ &= 5985 \end{align}$$Or using Pascal’s Identity: $\binom{20}{3} + \binom{20}{4} = \binom{21}{4} = 5985$
Example 2: Sum of Coefficients (JEE Main)
If the sum of coefficients in $(1 + x)^n$ is 1024, find $n$.
Solution: Sum of coefficients = $2^n = 1024 = 2^{10}$
Therefore, $n = 10$
Example 3: Odd/Even Position Sum (JEE Advanced)
In the expansion of $(1 + x)^{20}$, find the sum of coefficients of even powers of $x$.
Solution: Sum of coefficients at even powers = $2^{n-1} = 2^{20-1} = 2^{19} = 524288$
Example 4: Using Ratio Property (JEE Advanced)
In the expansion of $(1 + x)^n$, if the coefficient of 5th term is equal to the coefficient of 8th term, find $n$.
Solution: 5th term: $T_5 = \binom{n}{4}x^4$, coefficient = $\binom{n}{4}$
8th term: $T_8 = \binom{n}{7}x^7$, coefficient = $\binom{n}{7}$
Given: $\binom{n}{4} = \binom{n}{7}$
Using symmetry: $\binom{n}{4} = \binom{n}{n-4}$
So: $\binom{n}{7} = \binom{n}{n-4}$
Therefore: $7 = n - 4 \Rightarrow n = 11$
Example 5: Vandermonde’s Identity (JEE Advanced)
Prove: $\binom{2n}{0}\binom{2n}{2n} + \binom{2n}{1}\binom{2n}{2n-1} + \cdots + \binom{2n}{2n}\binom{2n}{0} = \binom{4n}{2n}$
Solution: Using Vandermonde’s Identity with $m = n = 2n$ and $k = 2n$:
$$\sum_{r=0}^{2n} \binom{2n}{r}\binom{2n}{2n-r} = \binom{4n}{2n}$$But $\binom{2n}{2n-r} = \binom{2n}{r}$ (symmetry)
Therefore: $\sum_{r=0}^{2n} \binom{2n}{r}\binom{2n}{r} = \binom{4n}{2n}$
Hence proved.
Practice Problems
Level 1: JEE Main Basics
Problem 1.1: Evaluate $\binom{15}{12} + \binom{15}{13}$ using Pascal’s identity.
Solution
Using Pascal’s Identity: $\binom{n-1}{r-1} + \binom{n-1}{r} = \binom{n}{r}$
Here: $\binom{15}{12} + \binom{15}{13} = \binom{16}{13}$
Using symmetry: $\binom{16}{13} = \binom{16}{3} = \frac{16 \times 15 \times 14}{6} = 560$
Problem 1.2: Find the sum of all binomial coefficients in $(1 + 2x)^{10}$.
Solution
Put $x = 1$: $(1 + 2)^{10} = 3^{10} = 59049$
Note: This gives sum of all coefficients including the constant multipliers.
For just binomial coefficients $\binom{10}{r}$: Sum = $2^{10} = 1024$
Problem 1.3: If $\binom{n}{8} = \binom{n}{6}$, find $n$.
Solution
Using symmetry: $\binom{n}{r} = \binom{n}{n-r}$
So: $\binom{n}{8} = \binom{n}{n-8}$
Given: $\binom{n}{8} = \binom{n}{6}$
Therefore: $6 = n - 8 \Rightarrow n = 14$
Level 2: JEE Main Advanced
Problem 2.1: Prove that $\binom{n}{1} + 2\binom{n}{2} + 3\binom{n}{3} + \cdots + n\binom{n}{n} = n \cdot 2^{n-1}$.
Solution
We know: $r\binom{n}{r} = n\binom{n-1}{r-1}$
So: $\sum_{r=1}^{n} r\binom{n}{r} = n\sum_{r=1}^{n}\binom{n-1}{r-1}$
Let $k = r-1$, then when $r = 1, k = 0$ and when $r = n, k = n-1$:
$= n\sum_{k=0}^{n-1}\binom{n-1}{k} = n \cdot 2^{n-1}$
Problem 2.2: Find the value of $\binom{10}{0}\binom{10}{10} + \binom{10}{1}\binom{10}{9} + \cdots + \binom{10}{10}\binom{10}{0}$.
Solution
Using Property 7: $\sum_{r=0}^{n}\binom{n}{r}^2 = \binom{2n}{n}$
Here $n = 10$:
$\sum_{r=0}^{10}\binom{10}{r}\binom{10}{10-r} = \sum_{r=0}^{10}\binom{10}{r}^2 = \binom{20}{10}$
$= \frac{20!}{10! \cdot 10!} = 184756$
Problem 2.3: If $\binom{n}{r-1} = 36$, $\binom{n}{r} = 84$, and $\binom{n}{r+1} = 126$, find $n$ and $r$.
Solution
Using $\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r}$:
$\frac{84}{36} = \frac{n-r+1}{r} \Rightarrow \frac{7}{3} = \frac{n-r+1}{r}$
$\Rightarrow 7r = 3(n-r+1) \Rightarrow 7r = 3n - 3r + 3 \Rightarrow 10r = 3n + 3$ … (1)
Similarly: $\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1}$:
$\frac{126}{84} = \frac{n-r}{r+1} \Rightarrow \frac{3}{2} = \frac{n-r}{r+1}$
$\Rightarrow 3(r+1) = 2(n-r) \Rightarrow 3r + 3 = 2n - 2r \Rightarrow 5r = 2n - 3$ … (2)
From (1): $n = \frac{10r - 3}{3}$
Substitute in (2): $5r = 2 \cdot \frac{10r - 3}{3} - 3$
$15r = 20r - 6 - 9 \Rightarrow 5r = 15 \Rightarrow r = 3$
From (2): $n = \frac{5(3) + 3}{2} = 9$
Therefore: $n = 9, r = 3$
Level 3: JEE Advanced
Problem 3.1: Prove that $\sum_{r=0}^{n} \frac{\binom{n}{r}}{r+1} = \frac{2^{n+1} - 1}{n+1}$.
Solution
We know: $(1+x)^n = \sum_{r=0}^{n}\binom{n}{r}x^r$
Integrate both sides from 0 to 1:
$\int_0^1 (1+x)^n dx = \sum_{r=0}^{n}\binom{n}{r}\int_0^1 x^r dx$
$\left[\frac{(1+x)^{n+1}}{n+1}\right]_0^1 = \sum_{r=0}^{n}\binom{n}{r}\left[\frac{x^{r+1}}{r+1}\right]_0^1$
$\frac{2^{n+1} - 1}{n+1} = \sum_{r=0}^{n}\frac{\binom{n}{r}}{r+1}$
Problem 3.2: Show that $\binom{n}{0}^2 - \binom{n}{1}^2 + \binom{n}{2}^2 - \cdots + (-1)^n\binom{n}{n}^2 = \begin{cases}0 & \text{if } n \text{ is odd} \\ (-1)^{n/2}\binom{n}{n/2} & \text{if } n \text{ is even}\end{cases}$
Solution
Consider $(1+x)^n(1-x)^n = (1-x^2)^n$
Coefficient of $x^n$ on LHS:
$\sum_{r=0}^{n}(-1)^{n-r}\binom{n}{r}\binom{n}{n-r} = (-1)^n\sum_{r=0}^{n}(-1)^r\binom{n}{r}^2$
Coefficient of $x^n$ on RHS from $(1-x^2)^n = \sum_{k=0}^{n}\binom{n}{k}(-x^2)^k$:
For $x^n$: need $2k = n$, so $k = n/2$ (only if $n$ is even)
If $n$ odd: coefficient = 0
If $n$ even: coefficient = $\binom{n}{n/2}(-1)^{n/2}$
Therefore: $(-1)^n\sum_{r=0}^{n}(-1)^r\binom{n}{r}^2 = \begin{cases}0 & n \text{ odd} \\ (-1)^{n/2}\binom{n}{n/2} & n \text{ even}\end{cases}$
Hence: $\sum_{r=0}^{n}(-1)^r\binom{n}{r}^2 = \begin{cases}0 & n \text{ odd} \\ (-1)^{n/2}\binom{n}{n/2} & n \text{ even}\end{cases}$
Problem 3.3: Prove: $\sum_{r=0}^{n} r^2 \binom{n}{r} = n(n+1)2^{n-2}$.
Solution
We know: $r\binom{n}{r} = n\binom{n-1}{r-1}$
So: $r^2\binom{n}{r} = rn\binom{n-1}{r-1} = n(r-1+1)\binom{n-1}{r-1}$
$= n(r-1)\binom{n-1}{r-1} + n\binom{n-1}{r-1}$
$= n(n-1)\binom{n-2}{r-2} + n\binom{n-1}{r-1}$
Therefore: $\sum_{r=0}^{n}r^2\binom{n}{r} = n(n-1)\sum_{r=0}^{n}\binom{n-2}{r-2} + n\sum_{r=0}^{n}\binom{n-1}{r-1}$
$= n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1}$
$= n \cdot 2^{n-2}[(n-1) + 2]$
$= n(n+1)2^{n-2}$
Important Identities Summary
| Identity | Formula |
|---|---|
| Symmetry | $\binom{n}{r} = \binom{n}{n-r}$ |
| Pascal’s | $\binom{n}{r} = \binom{n-1}{r-1} + \binom{n-1}{r}$ |
| Sum of all | $\sum_{r=0}^{n}\binom{n}{r} = 2^n$ |
| Odd/Even sum | $\sum_{r=0,2,4,\ldots}\binom{n}{r} = 2^{n-1}$ |
| Alternating sum | $\sum_{r=0}^{n}(-1)^r\binom{n}{r} = 0$ |
| Squares sum | $\sum_{r=0}^{n}\binom{n}{r}^2 = \binom{2n}{n}$ |
| Weighted sum | $\sum_{r=0}^{n}r\binom{n}{r} = n \cdot 2^{n-1}$ |
| Vandermonde | $\sum_{r=0}^{k}\binom{m}{r}\binom{n}{k-r} = \binom{m+n}{k}$ |
Cross-References
- Binomial Expansion: Foundation → Binomial Expansion
- General Term: Applications of coefficients → General Term
- Combinatorics: Counting principles → Combinatorics
- Sequences: Pascal’s triangle as sequence → Sequences
Quick Revision Checklist
- Symmetry property mastered
- Pascal’s identity understood
- Can calculate sum of all coefficients
- Know odd/even position sums
- Memorized ratio of consecutive coefficients
- Understand Vandermonde’s identity
- Practice with weighted sums
Last updated: October 22, 2025
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