Find the value of $r$, if the coefficients of $(2r + 4)^{\text{th}}$ and $(r - 2)^{\text{th}}$ terms in the expansion of $(1 + x)^{18}$ are equal.

Find the value of r, if the coefficients of (2r + – 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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The problem uses the property that binomial coefficients in $(1+x)^{18}$ are symmetric. Equating the coefficients of the $(2r+4)$ th and $(r-2)$ th terms gives a simple equation in $r$ , yielding $r = 6$ .

We are expanding $(1 + x)^{18}$ . The general term in the binomial expansion $(1 + x)^n$ is given by $T_{k+1} = \binom{n}{k} x^k$ , where the term number is $k+1$ (since the first term corresponds to $k=0$ ). So the coefficient of the $m$ th term is $\binom{n}{m-1}$ .

Here $n = 18$ . The coefficient of the $(2r+4)$ th term is $\binom{18}{(2r+4)-1} = \binom{18}{2r+3}$ .

The coefficient of the $(r-2)$ th term is $\binom{18}{(r-2)-1} = \binom{18}{r-3}$ .

We are told these coefficients are equal:

\binom{18}{2r+3} = \binom{18}{r-3}.

Now, the key property: binomial coefficients are symmetric. For any $n$ , $\binom{n}{k} = \binom{n}{n-k}$ . So two binomial coefficients with the same $n$ are equal either when the lower indices are equal, or when they are complementary (i.e., add up to $n$ ).

Watch out

A common mistake is to only set the lower indices equal and forget the complementary case. Always check both possibilities.

So we have two cases:

Case 1: $2r + 3 = r - 3$

This gives $2r - r = -3 - 3 \implies r = -6$ .

But $r$ must be such that the term numbers are positive integers. The $(r-2)$ th term requires $r-2 \ge 1$ , so $r \ge 3$ . $r = -6$ is invalid. Discard. …

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Q.Find the value of rrr, if the coefficients of (2r+4)th(2r + 4)^{\text{th}}(2r+4)th and (r−2)th(r - 2)^{\text{th}}(r−2)th terms in the expansion of (1+x)18(1 + x)^{18}(1+x)18 are equal.

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Q.Find the value of $r$ , if the coefficients of $(2r + 4)^{\text{th}}$ and $(r - 2)^{\text{th}}$ terms in the expansion of $(1 + x)^{18}$ are equal.

Unlock everything free for 14 days