Properties of Binomial Coefficients

Complete guide to binomial coefficient properties, identities, and applications in JEE problems

8 min read

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

IdentityFormula
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

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