$$ \text{If } \sum_{r=0}^5 \frac{^{11}C_{2r+1}}{2r+2}=\frac{m}{n},gcd(m, n)=1, \text{ then }m - n \text{ is equal to } $$ _______

We need to find $$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2}$$.

Using the identity: $$\frac{{}^{n}C_{k}}{k+1} = \frac{{}^{n+1}C_{k+1}}{n+1}$$

Here $$n = 11$$ and $$k = 2r+1$$, so:

$$\frac{{}^{11}C_{2r+1}}{2r+2} = \frac{{}^{12}C_{2r+2}}{12}$$

Therefore:

$$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{1}{12} \sum_{r=0}^{5} {}^{12}C_{2r+2}$$

The sum $$\sum_{r=0}^{5} {}^{12}C_{2r+2} = {}^{12}C_2 + {}^{12}C_4 + {}^{12}C_6 + {}^{12}C_8 + {}^{12}C_{10} + {}^{12}C_{12}$$

We know that the sum of even-indexed binomial coefficients equals: $$\sum_{k=0}^{6} {}^{12}C_{2k} = 2^{11} = 2048$$

This sum includes $${}^{12}C_0 = 1$$. So:

$$\sum_{r=0}^{5} {}^{12}C_{2r+2} = 2048 - {}^{12}C_0 = 2048 - 1 = 2047$$

Therefore:

$$\sum_{r=0}^{5} \frac{{}^{11}C_{2r+1}}{2r+2} = \frac{2047}{12}$$

Since $$2047 = 23 \times 89$$ and $$12 = 2^2 \times 3$$, they share no common factors, so $$\gcd(2047, 12) = 1$$.

Thus $$m = 2047$$ and $$n = 12$$.

$$m - n = 2047 - 12 = 2035$$