A committee of four persons is to be selected from 8 computer engineers and 3 mechanical engineers so as to include at least one mechanical engineers. The possible no. of ways are?

The correct answer is 260.

Here's how:

Total number of persons selected 11C4=330

Number of ways of selecting ONLY computer engineers and NO mechanical engineer is 8C4 = 70

Now, the condition CLEARLY says AT LEAST ONE mechanical engineer. So we just need to subtract ONLY computer engineers from Total engineers. 

Therefore, 330-70=260

The total number of ways of selecting 4 persons from 11 engineers is 11C4 = 330

The total number of ways of selecting 4 persons from 8 computer engineers (and no mechanical engineers) is 8C4 = 70

The total number of ways of selecting 3 persons from 8 computer engineers (and exactly one mechanical engineer from 4) is 8C3 * 3  = 168

So, the required number is 330 - 70 - 168 = 92

You can find good shortcuts of permutations and combinations for CAT here.

1 case:- If 1 Mec E selected
8C3*3C1
56×3 = 168
2:- 2 Mec E selected
8C2*3C2
28×3 = 84
3:- 3 Mec E selected
8C1*3C3
8×1=8
168+84+8=260