Q.If ${}^{n}C_9 = {}^{n}C_8$, find ${}^{n}C_{17}$.

Showing the 12 most recent of 17 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $^nC_8 = {}^nC_2$ then the value of $^nC_2$ is

   (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:

   $n = 8 + 2 = 10$

   Now compute:

   ${}^{10}C_2 = \dfrac{10!}{2!\,8!} = \dfrac{10\times9}{2} = 45$

   ✓Final answer

   (c) 45.

2. 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:

   $^{10}C_3 = \frac{10!}{3!\,7!} = \frac{10\times9\times8}{3\times2\times1} = \frac{720}{6} = 120$

   ✓Final answer

   $^nC_3 = 120$ — option (c).

3. CBSE 2026Set 1A1 markMCQ
   Q.If $^{12}C_{s+1} = {}^{12}C_{2s-5}$, then the value of $s$ is -

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

   Case 2: $(s+1)+(2s-5)=12 \Rightarrow 3s-4=12 \Rightarrow s=\tfrac{16}{3}$, not an integer, so rejected.

   ✓Final answer

   $s=6$, i.e. option (4).

4. CBSE 2025Set ANNUAL1 markMCQ
   Q.If ${}^nC_8 = {}^nC_6$ then the value of ${}^nC_3$ is

   (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:

   ${}^{14}C_3 = \frac{14!}{3!\,11!} = \frac{14\times13\times12}{3\times2\times1} = \frac{2184}{6} = 364$

   ✓Final answer

   (c) 364

5. 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 $^nC_6 = {}^nC_4$.

   The identity $^nC_r = {}^nC_{n-r}$ means $^nC_6 = {}^nC_4$ happens when either $6=4$ (impossible) or $6 = n-4$.

   $6 = n - 4 \Rightarrow n = 10$

   ✓Final answer

   $n = 10$.

6. CBSE 2024Set ANNUAL1 markMCQ
   Q.If ${}^{n}C_{12} = {}^{n}C_{8}$, then ${}^{n}C_{18} =$

   (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:

   $n = 12 + 8 = 20$

   Now compute $^{20}C_{18}$. Using the symmetry property again (since $18$ is close to $20$, it's easier to compute the complement):

   $^{20}C_{18} = {}^{20}C_{20-18} = {}^{20}C_2 = \frac{20 \times 19}{2 \times 1} = \frac{380}{2} = 190$

   ✓Final answer

   (b) $190$.

7. CBSE 2023Set ANNUAL1 markMCQ
   Q.If $^{n}C_{5} = {}^{n}C_{7}$, then $n =$

   (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:

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

   Given $^nC_5 = {}^nC_7$ with $5\ne7$, this can only hold if $7=n-5$ (the complementary pairing), so:

   $n = 5+7 = 12$

   ✓Final answer

   (c) 12.

8. CBSE 2023Set ANNUAL1 markMCQ
   Q.If ${}^{n}C_7 = {}^{n}C_5$, then the value of $n$ is

   (a) 2

   (b) 10

   (c) 12

   (d) 35

   ›Reveal solutionSolution

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

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

   Given ${}^nC_7={}^nC_5$, matching the lower indices gives $n-7=5$, so $n=12$.

   Check: ${}^{12}C_7={}^{12}C_5=792$. ✓

   ✓Final answer

   (iii) 12

9. 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 $r + s = n$.

   Here $r = 6$, $s = 8$, which are different, so:

   $n = r + s = 6 + 8 = 14$

   ✓Final answer

   $n = 14$.

10. CBSE 2022Set ANNUAL1 mark
    Q.If $^{n}C_8 = {}^{n}C_2$, find $^{n}C_2$.

    ›Reveal solutionSolution

    $^nC_2=45$.

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

    $n=8+2=10.$

    Then $^{10}C_2 = \dfrac{10\times9}{2}=45$.

    ✓Final answer

    $n=10$ and $^nC_2 = 45$.

11. CBSE 2022Set ANNUAL1 markMCQ
    Q.Fill in the blank with the correct option: if ${}^{n}C_{12} = {}^{n}C_{8}$, then $n = \_\_\_\_$

    (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 ${}^nC_r = {}^nC_{n-r}$).

    Here $r=12$ and $s=8$ are different, so we use $r+s=n$: $n = 12+8 = 20$.

    Check: ${}^{20}C_{12} = {}^{20}C_{20-12} = {}^{20}C_8$. ✓

    ✓Final answer

    $n = 20$.

12. CBSE 2022Set ANNUAL1 markMCQ
    Q.If $^nC_{12} = {}^nC_8$, then n is

    (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 $n=20$.

    A standard CBSE/NCERT Class 11 Permutations and Combinations identity.

    ✓Final answer

    (a) 20.