A committee of 5 people must be selected from accountants and 8 educators in how many ways can a selection done if there are at least 3 the committee?

We can select a committee of 5 people from 13 people in \( {13 \choose 5} = 1287\) ways.

However, we need to subtract the number of ways in which we can select a committee with fewer than 3 accountants.

There are \( {8 \choose 5} = 56\) ways to select a committee with 0 accountants.

There are \( {8 \choose 4} \times {5 \choose 1} = 140\) ways to select a committee with 1 accountant.

There are \( {8 \choose 3} \times {5 \choose 2} = 280\) ways to select a committee with 2 accountants.

So, the number of ways to select a committee with at least 3 accountants is:

$$1287 - 56 - 140 - 280 = 811$$