Out of 4 architects, 6 engineers and 9n electricians, how many ways can a group of 6 people be formed to attend a forum if there are at least 3 electricians?
To find the number of ways to form a group of 6 people with at least 3 electricians, we can use the concept of combinations.
First, let's consider the scenario where all 6 people in the group are electricians. Since there are 9 electricians available, we can select 6 of them in 9C6 ways (read as "9 choose 6").
Next, we need to consider the cases where 4 electricians and 2 people from other professions are selected. We can choose 4 electricians from the available 9 in 9C4 ways. Likewise, we can choose 2 people from the remaining architects and engineers in (4+6)C2 ways.
Finally, we need to consider the cases where 3 electricians, one architect, and two engineers are selected. We can choose 3 electricians from the available 9 in 9C3 ways. Similarly, we can choose one architect from the available 4 in 4C1 ways and two engineers from the available 6 in 6C2 ways.
To find the total number of ways, we can sum up the three cases:
Total ways = 9C6 + 9C4 * (4+6)C2 + 9C3 * 4C1 * 6C2
Now we can calculate the answer:
Total ways = (9! / (6! * (9-6)!)) + (9! / (4! * (9-4)!)) * (10! / (2! * (10-2)!)) + (9! / (3! * (9-3)!)) * (4! / (1! * (4-1)!)) * (6! / (2! * (6-2)!))
Simplifying the above expression will give us the answer to the problem.