Q.${}^nC_r + {}^nC_{r-1} = ?$

Showing the 12 most recent of 21 on this concept.

1. 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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2. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^{2025}C_{2025} =$

   (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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3. CBSE 2025Set ANNUAL1 markMCQ
   Q.$^{2025}C_1 =$

   (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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4. 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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5. CBSE 2025Set ANNUAL1 markMCQ
   Q.${}^nC_r + {}^nC_{r-1} = ?$

   (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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6. 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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7. CBSE 2025Set ANNUAL1 markMCQ
   Q.Match the column: Column A entry 'Value of ${}^5C_2$' — find the matching value from Column B.

   (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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8. 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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9. CBSE 2024Set ANNUAL1 markMCQ
   Q.If ${}^{n}P_{r} = 720$, ${}^{n}C_{r} = 120$, then $r = ?$

   (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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10. CBSE 2024Set ANNUAL1 markMCQ
    Q.If $n = 5$ and $r = 3$, then the value of ${}^{n}C_{r}$ is:

    (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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11. CBSE 2024Set hz1 markMCQ
    Q.If $n = 10$, $r = 3$, then value of ${}^{n}C_{r}$ is:

    (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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12. CBSE 2024Set ANNUAL1 markMCQ
    Q.The value of $^{n}C_{n}$ is:

    (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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