If ${}^{n}C_{12} = {}^{n}C_{8}$, then $n$ is equal to (A) $20$ (B) $12$ (C) $6$ (D) $30$

If ⁿC₁₂ = ⁿC₈, then n is equal to (A) 20 (B) 12 (C) 6 (D) 30

NCERT Exemplar · Q26

{}^{n}C_{12} = {}^{n}C_{8}

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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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This problem uses the symmetry property of combinations, ${}^{n}C_r = {}^{n}C_{n-r}$ , to find $n$ . Given ${}^{n}C_{12} = {}^{n}C_8$ , we deduce that $n = 12 + 8 = \boxed{20}$ .

When we deal with combinations, ${}^{n}C_r$ , we are counting the number of ways to select $r$ items from a set of $n$ distinct items, where the order of selection does not matter. A fundamental property of combinations, often called the symmetry property, is key to solving this problem.

The intuition behind this property is quite simple:

Imagine you have $n$ distinct items. If you choose $r$ items to take, you are simultaneously choosing $n-r$ items to leave behind. The number of ways to choose which $r$ items to take must be exactly the same as the number of ways to choose which $n-r$ items to leave behind. This is because each selection of $r$ items uniquely determines the set of $n-r$ items not chosen, and vice-versa.

{}^{n}C_r = {}^{n}C_{n-r}

Let's apply this understanding to the given problem.

Identify the given equation:

We are given the equation ${}^{n}C_{12} = {}^{n}C_8$ .

Here, we have two combination expressions that are equal, both involving the same total number of items, $n$ .
Apply the combination symmetry property:

According to the property, if ${}^{n}C_a = {}^{n}C_b$ , there are two possibilities:
- Possibility 1: The number of items being chosen is the same, i.e., $a = b$ .
- Possibility 2: The sum of the number of items being chosen is equal to the total number of items, i.e., $a + b = n$ .
Evaluate Possibility 1 ( $a=b$ ):

In our equation, $a=12$ and $b=8$ . …

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Q.If nC12=nC8{}^{n}C_{12} = {}^{n}C_{8}nC12​=nC8​, then nnn is equal to (A) 202020 (B) 121212 (C) 666 (D) 303030

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Q.If ${}^{n}C_{12} = {}^{n}C_{8}$ , then $n$ is equal to
(A) $20$
(B) $12$
(C) $6$
(D) $30$

Unlock everything free for 14 days