Q92 / 180 · COMEDK 2023 (Morning Shift) · Mathematics
$\text { Find }{ }^n C_{21} \text {, if }{ }^n C_{10}={ }^n C_{12}$
- A21
- B2
- Correct answerC22
- D1
Solution
The property of combinations states that if $^{n}C_{a} = ^{n}C_{b}$, then either $a = b$ or $a + b = n$.
Given $^{n}C_{10} = ^{n}C_{12}$, since $10 \neq 12$, it must be that $10 + 12 = n$.
Therefore, $n = 22$.
We are asked to find $^{n}C_{21}$, which is $^{22}C_{21}$.
Using the property $^{n}C_{r} = ^{n}C_{n-r}$, we have $^{22}C_{21} = ^{22}C_{22-21} = ^{22}C_{1}$.
Since $^{n}C_{1} = n$, we get $^{22}C_{1} = 22$.
Answer: 22
Given $^{n}C_{10} = ^{n}C_{12}$, since $10 \neq 12$, it must be that $10 + 12 = n$.
Therefore, $n = 22$.
We are asked to find $^{n}C_{21}$, which is $^{22}C_{21}$.
Using the property $^{n}C_{r} = ^{n}C_{n-r}$, we have $^{22}C_{21} = ^{22}C_{22-21} = ^{22}C_{1}$.
Since $^{n}C_{1} = n$, we get $^{22}C_{1} = 22$.
Answer: 22
Hence, the correct answer is option: 'C'.