If $$\frac{^{11}C_1}{2} + \frac{^{11}C_2}{3} + \ldots + \frac{^{11}C_9}{10} = \frac{n}{m}$$ with $$\gcd(n, m) = 1$$, then $$n + m$$ is equal to _______

We need to evaluate $$\sum_{r=1}^{9}\frac{^{11}C_r}{r+1}$$ and express it as $$\frac{n}{m}$$ with $$\gcd(n,m) = 1$$, then find $$n + m$$.

Use the identity $$\frac{\binom{n}{r}}{r+1} = \frac{1}{n+1}\binom{n+1}{r+1}$$.

This identity comes from:

$$\frac{\binom{n}{r}}{r+1} = \frac{n!}{r!(n-r)!(r+1)} = \frac{1}{n+1} \cdot \frac{(n+1)!}{(r+1)!(n-r)!} = \frac{1}{n+1}\binom{n+1}{r+1}$$

With $$n = 11$$:

$$\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\binom{12}{r+1}$$

Substitute and evaluate the sum.

$$\sum_{r=1}^{9}\frac{\binom{11}{r}}{r+1} = \frac{1}{12}\sum_{r=1}^{9}\binom{12}{r+1} = \frac{1}{12}\sum_{k=2}^{10}\binom{12}{k}$$

where $$k = r + 1$$.

Use the binomial sum.

We know: $$\sum_{k=0}^{12}\binom{12}{k} = 2^{12} = 4096$$

$$\sum_{k=2}^{10}\binom{12}{k} = 4096 - \binom{12}{0} - \binom{12}{1} - \binom{12}{11} - \binom{12}{12}$$

$$= 4096 - 1 - 12 - 12 - 1 = 4070$$

Final calculation.

$$\frac{n}{m} = \frac{4070}{12} = \frac{2035}{6}$$

Check: $$\gcd(2035, 6)$$. $$2035 = 5 \times 407 = 5 \times 11 \times 37$$. $$6 = 2 \times 3$$. No common factors, so $$\gcd = 1$$.

$$n + m = 2035 + 6 = 2041$$

The answer is 2041.