Q.a) Determine the number of 5 card combinations out of a pack of 52 playing cards if there is exactly one ace in each combination.

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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