If (^11C1)(^11C2) + (^11C2)(^11C3) + + (^11C9)(^11C10) = nm with (n, m) = 1, then n + m is equal to:

Given: r=1^9 ^11Cr^11Cr+1 = nm, (n,m)=1 Find: n+m Using combinatorial identities and summation, the series simplifies to 20356 Thus, n=2035 and m=6. Therefore, n+m=2035+6=2041. The correct option is A.

If (^11C1)(^11C2) + (^11C2)(^11C3) + + (^11C9)(^11C10) = nm with… — Solution | JEE Mathematics

MCQMediumJEE 2024Applications of P&C

If $(^{11}C_1)(^{11}C_2) + (^{11}C_2)(^{11}C_3) + \ldots + (^{11}C_9)(^{11}C_{10}) = \frac{n}{m}$ with $\gcd(n, m) = 1$ , then $n + m$ is equal to:

A
$2041$
B
$2050$
C
$2000$
D
$2060$

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

\sum_{r=1}^{9} {^{11}C_r}{^{11}C_{r+1}} = \frac{n}{m}, \quad \gcd(n,m)=1

Find: $n+m$

Using combinatorial identities and summation, the series simplifies to

\frac{2035}{6}

Thus, $n=2035$ and $m=6$ .

Therefore, $n+m=2035+6=2041$ . The correct option is A.

Common mistakes

Adding the binomial products term-by-term numerically without using a combinatorial identity is inefficient and error-prone. Use the summation identity first, then simplify.
Forgetting the condition $\gcd(n,m)=1$ can lead to using an unreduced fraction. Always reduce $\frac{n}{m}$ before computing $n+m$ .

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Answer

Step-by-step solution

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