The value of $^{n}C_{n}$ is: (a) 0 (b) undefined (c) 1 (d) 2

The value of ⁿCₙ is: (a) 0 (b) undefined (c) 1 (d) 2

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^{n}C_{n}

(a) 0

(b) undefined

(d) 2

Madhya Pradesh MpbseMP Board Higher Secondary (Class 11) 2024MCQ· 1mImportance★★★★★

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Concept understanding — Combinations Selection

Combinations: Choosing Without Ordering

Imagine you're picking a team of 3 players from a group of 5 friends: Alice, Bob, Charlie, Deepa, and Esha. The team {Alice, Bob, Charlie} is the same team as {Bob, Charlie, Alice} — the order you name them doesn't matter. What matters is which 3 people you pick.

That's the core idea of combinations: selection without regard to order.

The Intuition: Why Order Doesn't Matter

Let's contrast with permutations. If you were assigning positions — captain, vice-captain, treasurer — then {Alice as captain, Bob as vice-captain, Charlie as treasurer} is different from {Bob as captain, Alice as vice-captain, Charlie as treasurer}. Order matters there.

But for a plain team, a committee, a hand of cards, or a set of toppings on a pizza — order is irrelevant. You just care about which items are chosen.

Note

Key distinction: Permutations count arrangements (order matters). Combinations count selections (order doesn't matter).

From Permutations to Combinations

Suppose you want to choose 2 letters from {A, B, C}. If order mattered, you'd have these 6 permutations:

AB, BA, AC, CA, BC, CB

But if order doesn't matter, AB and BA are the same selection. So the distinct combinations are just:

{A, B}, {A, C}, {B, C} — only 3.

Notice the pattern: each combination of 2 items corresponds to $2! = 2$ permutations (because you can arrange those 2 items in 2 ways). So:

\text{Number of combinations} = \frac{\text{Number of permutations}}{r!}

Where $r$ is the number of items you're choosing.

The Precise Statement

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

This is read as "n choose r" and gives the number of ways to select $r$ distinct objects from a set of $n$ distinct objects, where order does not matter.

Conditions:

$n$ and $r$ are non-negative integers
$r \leq n$
The objects are distinct (no repetitions)

Why the Formula Works

Start with permutations of $r$ items from $n$ : $P(n, r) = \frac{n!}{(n-r)!}$ .

Each combination of $r$ items can be arranged in $r!$ different orders. So the number of combinations is the number of permutations divided by the number of ways to rearrange each selection:

\binom{n}{r} = \frac{P(n, r)}{r!} = \frac{n!}{r!(n-r)!}

Tip

A quick check: $\binom{n}{0} = 1$ (there's exactly one way to choose nothing), and $\binom{n}{n} = 1$ (one way to choose everything).

A Concrete Example

How many different 5-card hands can be dealt from a standard 52-card deck?

Here, $n = 52$ , $r = 5$ . The hand {A♠, K♥, Q♦, J♣, 10♠} is the same regardless of the order you receive the cards.

\binom{52}{5} = \frac{52!}{5! \cdot 47!} = \frac{52 \times 51 \times 50 \times 49 \times 48}{5 \times 4 \times 3 \times 2 \times 1} = 2,598,960

That's over 2.5 million possible hands — which is why poker is interesting. …

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$^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$ ; putting $r=n$ gives $^{n}C_{n}=1$ .

Step 1. Use the formula $^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$ .

…

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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 21 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. How many chords can be drawn through 21 points on a circle?$
(a) 420
(b) $^{21}P_2$
(c) 21
(d) 210
›Reveal solutionSolution

A chord is determined by an unordered pair of points, so count combinations (not permutations) of 21 points taken 2 at a time.

Every chord of a circle is uniquely determined by choosing 2 of the points on the circle (the order of choosing doesn't create a different chord — the chord from P to Q is the same as from Q to P). So this is a combinations problem:

${}^{21}C_2 = \dfrac{21!}{2!\,19!} = \dfrac{21\times20}{2} = 210$ …

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CBSE 2025Set ANNUAL1 markMCQ
$^{2025}C_{2025} =$
(a) 2025
(b) 0
(c) 1
(d) 2024
›Reveal solutionSolution

$^{2025}C_{2025} = 1$ .

The combination formula $^nC_r = \dfrac{n!}{r!(n-r)!}$ . For $r=n$ : $^nC_n = \dfrac{n!}{n!\,0!} = \dfrac{n!}{n!\times1} = 1$ (using $0!=1$ ), true for an …

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CBSE 2025Set ANNUAL1 markMCQ
$^{2025}C_1 =$
(a) 0
(b) 1
(c) 2025
(d) $2025!$
›Reveal solutionSolution

$^{2025}C_1 = 2025$ .

Using $^nC_r = \dfrac{n!}{r!(n-r)!}$ with $r=1$ : $^nC_1 = \dfrac{n!}{1!(n-1)!} = \dfrac{n(n-1)!}{(n-1)!} = n$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Relation between permutation and combination is$
(a) ${}^nC_r = \dfrac{{}^nP_r}{r!}$
(b) ${}^nC_r \times r! = 1$
(c) $\dfrac{{}^nC_r}{r!} = {}^nP_r$
(d) None of these
›Reveal solutionSolution

A permutation counts ordered selections, a combination counts unordered ones; each combination corresponds to $r!$ permutations (the orderings of its $r$ chosen items), so ${}^nC_r = {}^nP_r / r!$ .

${}^nP_r$ counts the number of ways to choose and arrange $r$ objects out of $n$ . Each unordered group of $r$ objects (a combination) can be arranged in $r!$ different orders (permutations).

So: …

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CBSE 2025Set ANNUAL1 markMCQ
${}^nC_r + {}^nC_{r-1} = ?$
(a) ${}^{n+1}C_r$
(b) ${}^{n-1}C_r$
(c) ${}^{n+1}C_{r+1}$
(d) None of these
›Reveal solutionSolution

${}^nC_r + {}^nC_{r-1} = {}^{n+1}C_r$ is the standard Pascal's-rule identity for combinations.

This is a fundamental combinatorial identity (Pascal's Rule), which can be proved algebraically:

${}^nC_r + {}^nC_{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$

Taking $\frac{n!}{r!(n-r+1)!}$ as the common structure and combining the terms simplifies (via the standard derivation) to:

$\frac{(n+1)!}{r!(n+1-r)!} = {}^{n+1}C_r$

…

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CBSE 2025Set ANNUAL1 markMCQ
$Q. How many different teams of 7 players can be chosen out of 10 players?$
(a) 720
(b) 120
(c) 70
(d) None of these
›Reveal solutionSolution

A team of 7 from 10 players is ${}^{10}C_7$ , which by symmetry equals ${}^{10}C_3$ .

…

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CBSE 2025Set ANNUAL1 markMCQ
${}^5C_2$
(a) 2
(b) 8
(c) $\dfrac{2}{3}$
(d) $\dfrac{2\tan x}{1 - \tan^2 x}$
(e) $\sin 2x$
(f) 10
(g) 20
(h) $\dfrac{2\tan x}{1 + \tan^2 x}$
(i) 4
›Reveal solutionSolution

$^5C_2 = \dfrac{5!}{2!\,3!} = 10$ , matching Column B option (f).

…

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CBSE 2024Set ANNUAL1 markMCQ
$Q. In how many ways a team of 3 boys and 4 girls can be selected from 5 boys and 4 girls?$
(a) ${}^{5}C_{3} \times {}^{4}C_{4}$
(b) ${}^{5}C_{3} \times {}^{4}C_{3}$
(c) ${}^{5}C_{4} \times {}^{4}C_{3}$
(d) None of these
›Reveal solutionSolution

When two independent selections must both happen, multiply the number of ways for each (fundamental principle of counting).

We need to select a team of 3 boys and 4 girls from a pool of 5 boys and 4 girls.

Choosing the boys: we need 3 boys out of 5 available, which can be done in $^5C_3$ ways.

…

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CBSE 2024Set ANNUAL1 markMCQ
${}^{n}P_{r} = 720$
(a) $r=3$
(b) $r=5$
(c) $r=4$
(d) None of these
›Reveal solutionSolution

Permutations and combinations are related by $^nP_r = {}^nC_r \times r!$ ; use this to solve for $r!$ directly.

Recall the relationship between permutations and combinations:

$^nP_r = {}^nC_r \times r!$

We are given $^nP_r = 720$ and $^nC_r = 120$ . Substituting:

$720 = 120 \times r!$ …

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CBSE 2024Set ANNUAL1 markMCQ
$n = 5$
(a) $10$
(b) $30$
(c) $10!$
(d) $15!$ .
›Reveal solutionSolution

${}^5C_3=10$ .

The combination formula is ${}^nC_r=\dfrac{n!}{r!(n-r)!}$ . With $n=5,\ r=3$ : …

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CBSE 2024Set hz1 markMCQ
$n = 10$
(a) 100
(b) 120
(c) 110
(d) 520
›Reveal solutionSolution

${}^{10}C_3 = \dfrac{10!}{3!\,7!} = 120$ .

The formula for combinations is

${}^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$

With $n = 10$ , $r = 3$ : …

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CBSE 2024Set ANNUAL1 markMCQ
$^{n}C_{n}$
(a) 0
(b) undefined
(c) 1
(d) 2
›Reveal solutionSolution

$^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$ ; putting $r=n$ gives $^{n}C_{n}=1$ .

Step 1. Use the formula $^{n}C_{r} = \dfrac{n!}{r!(n-r)!}$ .

…

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