If $^{n}C_{5} = {}^{n}C_{7}$, then $n =$ (a) 5 (b) 7 (c) 12 (d) 0

If ⁿC₅ = ⁿC₇, then n = (a) 5 (b) 7 (c) 12 (d) 0

Question of 130

^{n}C_{5} = {}^{n}C_{7}

(a) 5

(b) 7

(d) 0

Jharkhand JacJAC Intermediate Board (1st Year) 2023MCQ· 1mImportance★★★★★

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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 identity $^nC_r = {}^nC_{n-r}$ means two equal combination values (with different lower indices) must have indices summing to $n$ .

A key combinatorics identity is:

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

…

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 17 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$^nC_8 = {}^nC_2$
(a) 54
(b) 10
(c) 45
(d) None of these
›Reveal solutionSolution

Use the identity $^nC_r = {}^nC_s \implies r+s=n$ (for $r\ne s$ ) to find $n$ , then compute $^nC_2$ .

The combination identity states: if $^nC_r = {}^nC_s$ with $r\ne s$ , then $r + s = n$ . Here $r=8, s=2$ , so:

$n = 8 + 2 = 10$ …

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If ⁿC₈ = ⁿC₂, then the value of ⁿC₃ is:$
(a) 720
(b) 360
(c) 120
(d) None of these
›Reveal solutionSolution

nC8 = nC2 forces n = 10 (since nCr = nC(n-r) implies r + s = n); then compute 10C3 = 120.

Using the identity $^nC_r = {}^nC_s \iff r+s=n$ (or $r=s$ ), and since $8\ne 2$ , we must have:

$8 + 2 = n \implies n = 10$

Now: …

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CBSE 2026Set 1A1 markMCQ
$^{12}C_{s+1} = {}^{12}C_{2s-5}$
(1) $4$
(2) $8$
(3) $12$
(4) $6$
›Reveal solutionSolution

$^{12}C_{s+1}={}^{12}C_{2s-5}$ gives $s=6$ .

If $^{n}C_a={}^{n}C_b$ then either $a=b$ or $a+b=n$ .

Case 1: $s+1=2s-5 \Rightarrow s=6$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
${}^nC_8 = {}^nC_6$
(a) 350
(b) 346
(c) 364
(d) 580
›Reveal solutionSolution

Using ${}^nC_r = {}^nC_s \implies r+s=n$ (for $r\neq s$ ), $n=8+6=14$ . Then ${}^{14}C_3 = 364$ .

${}^nC_8 = {}^nC_6$

Since ${}^nC_r = {}^nC_{n-r}$ , having ${}^nC_8={}^nC_6$ with $8\ne6$ means $8=n-6$ , i.e.

$n = 8+6 = 14$

Now compute: …

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CBSE 2025Set ANNUAL1 mark
$Q. If ⁿC₆ = ⁿC₄, then the value of n is .............$
›Reveal solutionSolution

When two different combination values from the same $n$ are equal, their lower indices add up to $n$ .

Given $^nC_6 = {}^nC_4$ .

…

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CBSE 2024Set ANNUAL1 markMCQ
${}^{n}C_{12} = {}^{n}C_{8}$
(a) 66
(b) 190
(c) 120
(d) None of these
›Reveal solutionSolution

Use the symmetry property $^nC_r = {}^nC_{n-r}$ to first find $n$ , then compute the required combination.

We are given $^nC_{12} = {}^nC_8$ . Using the identity $^nC_r = {}^nC_{n-r}$ , this equality holds when either $12 = 8$ (false) or $12 = n - 8$ , giving:

$n = 12 + 8 = 20$

…

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CBSE 2023Set ANNUAL1 markMCQ
$^{n}C_{5} = {}^{n}C_{7}$
(a) 5
(b) 7
(c) 12
(d) 0
›Reveal solutionSolution

The identity $^nC_r = {}^nC_{n-r}$ means two equal combination values (with different lower indices) must have indices summing to $n$ .

A key combinatorics identity is:

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

…

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CBSE 2023Set ANNUAL1 markMCQ
${}^{n}C_7 = {}^{n}C_5$
(a) 2
(b) 10
(c) 12
(d) 35
›Reveal solutionSolution

Use the identity ${}^nC_r={}^nC_{n-r}$ .

By the identity ${}^nC_r={}^nC_{n-r}$ , ${}^nC_7={}^nC_{n-7}$ .

…

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CBSE 2022Set ANNUAL1 mark
$Q. If ⁿC₆ = ⁿC₈, then the value of n is ............ .$
›Reveal solutionSolution

Since $6\ne 8$ , the equal-combination property gives $n = 6 + 8 = 14$ .

For combinations, $^nC_r = {}^nC_s$ implies either $r = s$ , or $r + s = n$ .

…

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CBSE 2022Set ANNUAL1 mark
$^{n}C_8 = {}^{n}C_2$
›Reveal solutionSolution

$^nC_2=45$ .

Using $^nC_r={}^nC_s\Rightarrow r+s=n$ (for $r\ne s$ ): here $^nC_8={}^nC_2$ gives

$n=8+2=10.$

…

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CBSE 2022Set ANNUAL1 markMCQ
${}^{n}C_{12} = {}^{n}C_{8}$
(a) 11
(b) 14
(c) 20
›Reveal solutionSolution

${}^nC_{12}={}^nC_8$ implies $n=12+8=20$ .

For combinations, ${}^nC_r = {}^nC_s$ holds if either $r=s$ , or $r+s=n$ (since ${}^nC_r = {}^nC_{n-r}$ ).

…

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CBSE 2022Set ANNUAL1 markMCQ
$^nC_{12} = {}^nC_8$
(a) 20
(b) 12
(c) 6
(d) 8
›Reveal solutionSolution

$n=20$ .

If $^nC_r={}^nC_s$ with $r e s$ , then $r+s=n$ . Here $12+8=20$ , so $n=20$ .

…

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Q.If nC5=nC7^{n}C_{5} = {}^{n}C_{7}nC5​=nC7​, then n=n =n=

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