From a group of $7\text{men}$ and $6\text{women}$, five persons are to be selected to form a committee so that at least $3\text{men}$ are there in the committee. In how many ways can it be done?

To form a committee of $5$ members containing at least 3 men, we must consider all possible valid combinations. The constraint allows for exactly $3$, $4$, or $5$ men.
Let's break this down into three mutually exclusive cases using the combination formula $\left(\genfrac{}{}{0ex}{}{n}{r}\right)=\frac{n!}{r!(n-r)!}$:

* Case 1: Exactly 3 men and 2 women

$$\text{Ways}=\left(\genfrac{}{}{0ex}{}{7}{3}\right)\times \left(\genfrac{}{}{0ex}{}{6}{2}\right)$$

$$\left(\genfrac{}{}{0ex}{}{7}{3}\right)=\frac{7\times 6\times 5}{3\times 2\times 1}=35$$

$$\left(\genfrac{}{}{0ex}{}{6}{2}\right)=\frac{6\times 5}{2\times 1}=15$$

$$\text{Ways for Case 1}=35\times 15=525$$

* Case 2: Exactly 4 men and 1 woman

$$\text{Ways}=\left(\genfrac{}{}{0ex}{}{7}{4}\right)\times \left(\genfrac{}{}{0ex}{}{6}{1}\right)$$

$$\left(\genfrac{}{}{0ex}{}{7}{4}\right)=\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$$

$$\left(\genfrac{}{}{0ex}{}{6}{1}\right)=6$$

$$\text{Ways for Case 2}=35\times 6=210$$

* Case 3: Exactly 5 men and 0 women

$$\text{Ways}=\left(\genfrac{}{}{0ex}{}{7}{5}\right)\times \left(\genfrac{}{}{0ex}{}{6}{0}\right)$$

$$\left(\genfrac{}{}{0ex}{}{7}{5}\right)=\frac{7\times 6}{2\times 1}=21$$

$$\left(\genfrac{}{}{0ex}{}{6}{0}\right)=1$$

$$\text{Ways for Case 3}=21\times 1=21$$

* Total Number of Ways:
Summing up all individual combinations:

$$\text{Total Ways}=525+210+21=756$$