The coefficient of $a^{-6} b^4$ in the expansion of $\left(\dfrac{1}{a} - \dfrac{2b}{3}\right)^{10}$ is ______ .

The coefficient of a⁻⁶ b⁴ in the expansion of ((1) – 1st PUC Mathematics

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a^{-6} b^4

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Concept understanding — Binomial Theorem Expansion

The Problem That Started It All

Imagine you have to expand $(x + y)^2$ . That's easy: $x^2 + 2xy + y^2$ . Now try $(x + y)^3$ : $x^3 + 3x^2y + 3xy^2 + y^3$ . Still manageable.

But what about $(x + y)^{10}$ ? Or $(x + y)^{100}$ ? Multiplying it out term by term would take forever. There has to be a pattern — and there is.

The Binomial Theorem is the shortcut that tells you exactly what each term in the expansion of $(x + y)^n$ looks like, without ever having to multiply.

The Pattern You Already Know

Look at the expansions you already know:

Power	Expansion
$(x + y)^0$	$1$
$(x + y)^1$	$x + y$
$(x + y)^2$	$x^2 + 2xy + y^2$
$(x + y)^3$	$x^3 + 3x^2y + 3xy^2 + y^3$
$(x + y)^4$	$x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$

Notice three things:

The powers of $x$ decrease from $n$ down to $0$ , while the powers of $y$ increase from $0$ up to $n$ . In every term, the exponents add to $n$ .
The coefficients — 1, 4, 6, 4, 1 for $n=4$ — follow a famous pattern called Pascal's triangle.
The number of terms is always $n+1$ .

Note

Pascal's triangle: each number is the sum of the two numbers directly above it.

       1
      1 1
     1 2 1
    1 3 3 1
   1 4 6 4 1
  1 5 10 10 5 1

The Precise Statement

n

Let's break that down.

The symbol $\sum_{k=0}^{n}$ means "add up terms for $k = 0, 1, 2, \dots, n$ ". For each $k$ , the term is:

$\binom{n}{k}$ — the coefficient (read as "n choose k")
$x^{n-k}$ — $x$ raised to the power $n-k$
$y^k$ — $y$ raised to the power $k$

So for $n=4$ , the terms are:

$k$	$\binom{4}{k}$	$x^{4-k}$	$y^k$	Term
0	$\binom{4}{0}=1$	$x^4$	$y^0$	$1 \cdot x^4$
1	$\binom{4}{1}=4$	$x^3$	$y^1$	$4x^3y$
2	$\binom{4}{2}=6$	$x^2$	$y^2$	$6x^2y^2$
3	$\binom{4}{3}=4$	$x^1$	$y^3$	$4xy^3$
4	$\binom{4}{4}=1$	$x^0$	$y^4$	$1 \cdot y^4$

Add them up: $x^4 + 4x^3y + 6x^2y^2 + 4xy^3 + y^4$ . Exactly what we had.

Where Do Those Coefficients Come From?

The binomial coefficient $\binom{n}{k}$ counts how many ways you can choose $k$ items from a set of $n$ items. In the expansion, it counts how many ways you can pick $k$ copies of $y$ (and therefore $n-k$ copies of $x$ ) when multiplying $(x+y)$ by itself $n$ times.

Tip

To compute $\binom{n}{k}$ quickly: start at $n$ and multiply $k$ decreasing numbers, then divide by $k!$ .

Example: $\binom{7}{3} = \frac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1} = 35$ .

The General Term

The $k$ -th term (starting with $k=0$ ) in the expansion is:

\text{General term: } T_{k+1} = \binom{n}{k} x^{n-k} y^k

This is the most useful part for exams. If someone asks "find the 5th term in $(x+y)^{10}$ ", you set $k=4$ (because $T_{k+1}$ means $k=4$ gives the 5th term) and write:

T_5 = \binom{10}{4} x^{6} y^{4}

What If It's Not Just $x$ and $y$ ? …

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Take $k=4$ in $\binom{10}{k}\left(\frac1a\right)^{10-k}\left(-\frac{2b}{3}\right)^{k}$ to get $a^{-6}b^4$ ; the coefficient is $210\cdot\dfrac{16}{81}=\dfrac{1120}{27}$ .

The general term of $\left(\dfrac{1}{a}-\dfrac{2b}{3}\right)^{10}$ is

\binom{10}{k}\left(\frac{1}{a}\right)^{10-k}\left(-\frac{2b}{3}\right)^{k} =\binom{10}{k}\,\frac{(-2)^{k}}{3^{k}}\;a^{-(10-k)}\,b^{k}.

For $a^{-6}b^{4}$ we need $k=4$ (which also makes $-(10-k)=-6$ ). Then …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

KCET 2025Set A-11 markMCQ
$(2x+3)^{3n}$
›Reveal solutionSolution

Number of terms in $(a+b)^N$ is $N+1$ ; set $3n + 1 = 22$ and solve.

Step 1 — The concept: how many terms a binomial expansion has

By the binomial theorem,
$(a+b)^{N} = \sum_{r=0}^{N} {}^{N}C_{r}\, a^{N-r} b^{r}$
The index $r$ runs over $r = 0, 1, 2, \dots, N$ — that is $N + 1$ values, and each gives one distinct term. So:
$\text{Number of terms} = N + 1$
(The "+1" is simply because the count starts at $r=0$ , not $r=1$ — the commonest slip in this type of question.)

Step 2 — Apply it to the given expansion

For $(2x+3)^{3n}$ the exponent is $N = 3n$ . Note that the coefficients 2 and 3 are irrelevant — only the power governs the number of terms. Hence

$\text{Number of terms} = 3n + 1$ …
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KCET 2024Set A-11 markMCQ
$^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{45}C_3 + ^{45}C_4$
›Reveal solutionSolution

The sum telescopes using the Pascal identity $\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ , reducing to $\binom{50}{4}$ .

The problem asks for the value of

$^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{45}C_3 + ^{45}C_4$ .

At first glance, this is a sum of several binomial coefficients. The last term, $^{45}C_4$ , is the odd one out — it has a different bottom index. That’s the key: it pairs naturally with $^{45}C_3$ using Pascal’s identity.

Pascal’s identity says:

$\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}$ .

So $^{45}C_3 + ^{45}C_4 = ^{46}C_4$ .

Now the sum becomes:

$^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{46}C_3 + ^{46}C_4$ .

Again, $^{46}C_3 + ^{46}C_4 = ^{47}C_4$ .

So we have: $^{49}C_3 + ^{48}C_3 + ^{47}C_3 + ^{47}C_4$ .

Next, $^{47}C_3 + ^{47}C_4 = ^{48}C_4$ .

Now: $^{49}C_3 + ^{48}C_3 + ^{48}C_4$ .

Then $^{48}C_3 + ^{48}C_4 = ^{49}C_4$ .

So: $^{49}C_3 + ^{49}C_4$ .

Finally, $^{49}C_3 + ^{49}C_4 = ^{50}C_4$ .

Each step is just applying Pascal’s identity to the last two terms, collapsing the sum from right to left like a telescope. …
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COMEDK 2024Set 2024-E1 markMCQ
$\left(\frac{1}{x}+x \sin x\right)^{10}, \quad$
›Reveal solutionSolution
The problem uses the binomial expansion of $\left(\frac{1}{x}+x\sin x\right)^{10}$ ; the 6th term’s coefficient is given as $7\frac{7}{8}$ , which simplifies to $\frac{63}{8}$ . Equating the binomial coefficient times the powers of $x$ and $\sin x$ yields $\sin^5 x = \frac{1}{32}$ , so $\sin x = \frac12$ , giving $x = 30^\circ$ .

The key idea is that in a binomial expansion $(a+b)^n$ , the $r$ -th term (starting with $r=0$ for the first term) is $\binom{n}{r} a^{n-r} b^r$ . Here the 6th term corresponds to $r=5$ . The coefficient of that term is not just the binomial coefficient — it also includes the numerical factors from $a$ and $b$ . Since $a = \frac{1}{x}$ and $b = x \sin x$ , the powers of $x$ cancel, leaving only powers of $\sin x$ . The given coefficient then becomes an equation in $\sin x$ .

Identify the term number correctly In $(a+b)^{10}$ , the first term has $r=0$ , second term $r=1$ , …, so the 6th term has $r=5$ . The general term is

$T_{r+1} = \binom{10}{r} a^{10-r} b^r.$
For the 6th term ( $r=5$ ):
$T_6 = \binom{10}{5} a^{5} b^{5}.$

Substitute $a = \frac{1}{x}$ and $b = x \sin x$

$T_6 = \binom{10}{5} \left(\frac{1}{x}\right)^5 (x \sin x)^5 = \binom{10}{5} \frac{1}{x^5} \cdot x^5 \sin^5 x.$
The $x^5$ cancels completely:
$T_6 = \binom{10}{5} \sin^5 x.$

Compute the binomial coefficient

$\binom{10}{5} = \frac{10 \cdot 9 \cdot 8 \cdot 7 \cdot 6}{5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} = 252.$
So the coefficient of the 6th term is $252 \sin^5 x$ .

Set equal to the given coefficient The problem states this coefficient equals $7\frac{7}{8} = \frac{63}{8}$ .

$252 \sin^5 x = \frac{63}{8}.$
Divide both sides by 63:
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COMEDK 2021Set 2021-B1 markMCQ
$(x+a)^{53}+(x-a)^{53}$
›Reveal solutionSolution

There are $27$ terms. …

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Q.The coefficient of a−6b4a^{-6} b^4a−6b4 in the expansion of (1a−2b3)10\left(\dfrac{1}{a} - \dfrac{2b}{3}\right)^{10}(a1​−32b​)10 is ______ .

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Q.The coefficient of $a^{-6} b^4$ in the expansion of $\left(\dfrac{1}{a} - \dfrac{2b}{3}\right)^{10}$ is ______ .

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