The ratio of the coefficients of xp and xq in the expansion of (1 + x)p + q is ___ .

NCERT Exemplar · Q31

x^p

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Concept understanding — Combinations Symmetry Property

The Intuition: Two Ways to Choose

Imagine you have a group of 10 friends, and you need to pick 3 of them to form a committee. One way to think about this is: you are choosing the 3 people who will be on the committee. But there is another, equally valid way to think about it: you are rejecting the 7 people who will not be on the committee.

Choosing 3 to include is the same decision as choosing 7 to exclude. Every time you pick a set of 3, you automatically determine the set of 7 who are left out. There is a perfect one-to-one match between the two choices.

This is the heart of the symmetry property: the number of ways to choose $k$ items from $n$ is exactly the same as the number of ways to choose $n-k$ items from $n$ .

The Precise Statement

\binom{n}{k} = \binom{n}{n-k}

Where $\binom{n}{k}$ (read "n choose k") is the number of combinations — the number of distinct subsets of size $k$ you can pick from a set of $n$ distinct objects.

This holds for any non-negative integers $n$ and $k$ where $0 \leq k \leq n$ .

Why It Works (The Algebraic Proof)

The formula for combinations is:

\binom{n}{k} = \frac{n!}{k!(n-k)!}

Now compute $\binom{n}{n-k}$ :

\binom{n}{n-k} = \frac{n!}{(n-k)!(n-(n-k))!} = \frac{n!}{(n-k)!k!}

The denominator is just $k!(n-k)!$ written in a different order. Since multiplication is commutative, the two expressions are identical.

Note

The symmetry is purely algebraic, but the intuition is what makes it memorable: choosing $k$ to keep is the same as choosing $n-k$ to discard.

Special Cases That Make Sense

$k = 0$ : $\binom{n}{0} = 1$ (there is exactly one way to choose nothing). By symmetry, $\binom{n}{n} = 1$ (one way to choose everything). Both make sense — you either take nothing or take all.
$k = 1$ : $\binom{n}{1} = n$ . Symmetry gives $\binom{n}{n-1} = n$ . Choosing 1 person to include is the same as choosing $n-1$ people to exclude — there are $n$ choices in either case. …

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The key idea is to use the binomial theorem to find the coefficients and then apply the symmetry property of binomial coefficients. The ratio of the coefficients of $x^p$ and $x^q$ in $(1+x)^{p+q}$ is 1.

The problem asks for the ratio of coefficients of specific terms in a binomial expansion. The core concept here is the Binomial Theorem, specifically how to find the coefficient of a particular power of $x$ in the expansion of $(1+x)^n$ . A crucial property of binomial coefficients, their symmetry, will simplify the final ratio significantly.

The binomial expansion of $(1+x)^n$ is given by:

(1+x)^n = \binom{n}{0} + \binom{n}{1}x + \binom{n}{2}x^2 + \dots + \binom{n}{k}x^k + \dots + \binom{n}{n}x^n

From this, we see that the coefficient of $x^k$ in the expansion of $(1+x)^n$ is $\binom{n}{k}$ .

x^k

In this problem, the power of the binomial is $n = p+q$ . So, we are expanding $(1+x)^{p+q}$ .

Find the coefficient of $x^p$ :

Using the formula, the coefficient of $x^p$ in the expansion of $(1+x)^{p+q}$ is obtained by setting $k=p$ and $n=p+q$ .

So, the coefficient of $x^p$ is $\binom{p+q}{p}$ .
Find the coefficient of $x^q$ :

Similarly, the coefficient of $x^q$ in the expansion of $(1+x)^{p+q}$ is obtained by setting $k=q$ and $n=p+q$ .

So, the coefficient of $x^q$ is $\binom{p+q}{q}$ .
Form the ratio:

We need to find the ratio of the coefficient of $x^p$ to the coefficient of $x^q$ .

\text{Ratio} = \frac{\text{Coefficient of } x^p}{\text{Coefficient of } x^q} = \frac{\binom{p+q}{p}}{\binom{p+q}{q}}

Simplify the ratio using the symmetry property of binomial coefficients: …

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Q.The ratio of the coefficients of xpx^pxp and xqx^qxq in the expansion of (1+x)p+q(1 + x)^{p + q}(1+x)p+q is ______ .

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Q.The ratio of the coefficients of $x^p$ and $x^q$ in the expansion of $(1 + x)^{p + q}$ is ______ .

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