If $A$ and $B$ are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ respectively, then $\dfrac{A}{B}$ equals (A) $1$ (B) $2$ (C) $\dfrac{1}{2}$ (D) $\dfrac{1}{n}$

If A and B are coefficient of xⁿ in the expansions – Class 11 Mathematics

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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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Using the symmetry property of binomial coefficients, we recognize that $A = \binom{2n}{n}$ and $B = \binom{2n-1}{n}$ . The ratio simplifies by expressing $A$ in terms of $B$ through the Pascal's triangle identity, yielding $\dfrac{A}{B} = 2$ .

The heart of this problem lies in understanding how binomial coefficients behave and relate to one another. When we expand $(1+x)^m$ , the coefficient of $x^k$ is simply $\binom{m}{k}$ . But there's a beautiful symmetry at play here, and recognizing it transforms what looks like a calculation into an elegant observation.

The coefficient $A$ of $x^n$ in $(1+x)^{2n}$ is $\binom{2n}{n}$ —the middle term of the expansion. This is the largest binomial coefficient in that row of Pascal's triangle, sitting right at the center. Meanwhile, $B$ is the coefficient of $x^n$ in $(1+x)^{2n-1}$ , which is $\binom{2n-1}{n}$ .

Now here's the key insight: Pascal's triangle tells us that any entry is the sum of the two entries directly above it. Specifically:

\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}

But there's more. The binomial coefficients in any expansion satisfy the symmetry property $\binom{m}{k} = \binom{m}{m-k}$ . Applying this to the first term on the right:

\binom{2n-1}{n-1} = \binom{2n-1}{(2n-1)-(n-1)} = \binom{2n-1}{n}

This is remarkable: both terms in Pascal's identity are equal! Therefore:

\binom{2n}{n} = \binom{2n-1}{n} + \binom{2n-1}{n} = 2\binom{2n-1}{n}

Let me work through this systematically: …

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Q.If AAA and BBB are coefficient of xnx^nxn in the expansions of (1+x)2n(1 + x)^{2n}(1+x)2n and (1+x)2n−1(1 + x)^{2n - 1}(1+x)2n−1 respectively, then AB\dfrac{A}{B}BA​ equals (A) 111 (B) 222 (C) 12\dfrac{1}{2}21​ (D) 1n\dfrac{1}{n}n1​

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Q.If $A$ and $B$ are coefficient of $x^n$ in the expansions of $(1 + x)^{2n}$ and $(1 + x)^{2n - 1}$ respectively, then $\dfrac{A}{B}$ equals
(A) $1$
(B) $2$
(C) $\dfrac{1}{2}$
(D) $\dfrac{1}{n}$

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