The number of terms in the expansion of $(x + y + z)^n$ ______ .

The number of terms in the expansion of (x + y + z)ⁿ ___ .

NCERT Exemplar · Q26

(x + y + z)^n

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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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The expansion of $(x + y + z)^n$ is a sum of monomials of the form $x^a y^b z^c$ where $a + b + c = n$ ; counting these is a classic "stars and bars" problem. The number of terms is $\boxed{\frac{(n+2)!}{n! \cdot 2!}} = \frac{(n+1)(n+2)}{2}$ .

When we expand $(x + y + z)^n$ , we're multiplying $n$ copies of the trinomial together. Each term in the expansion comes from choosing one variable from each of the $n$ factors. The result is a monomial $x^a y^b z^c$ where the exponents $a$ , $b$ , and $c$ count how many times we picked $x$ , $y$ , and $z$ respectively. Since we make exactly $n$ choices in total, we must have $a + b + c = n$ with $a, b, c \geq 0$ .

The question reduces to: how many non-negative integer solutions does $a + b + c = n$ have?

This is the combinatorial problem of distributing $n$ identical objects into $3$ distinct bins. The standard technique is "stars and bars."

Counting with stars and bars

Visualize the problem: Imagine $n$ stars in a row: $\star \star \star \cdots \star$ . We need to partition them into three groups (for $x$ , $y$ , and $z$ ). We do this by inserting $2$ dividers (bars) among the stars.
Total positions: We have $n$ stars and need to place $2$ bars. Think of it as arranging $n + 2$ objects in a line: $n$ indistinguishable stars and $2$ indistinguishable bars.
Choose positions for the bars: The number of ways to choose $2$ positions out of $n + 2$ total positions for the bars is $\binom{n+2}{2}$ .
Compute the binomial coefficient:

\binom{n+2}{2} = \frac{(n+2)!}{2! \cdot n!} = \frac{(n+2)(n+1)}{2}

Tip

The general formula for distributing $n$ identical items into $k$ distinct bins is $\binom{n+k-1}{k-1}$ . Here $k = 3$ , so we get $\binom{n+2}{2}$ . …

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Q.The number of terms in the expansion of (x+y+z)n(x + y + z)^n(x+y+z)n ______ .

Counting with stars and bars

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Q.The number of terms in the expansion of $(x + y + z)^n$ ______ .

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