A committee of 6 is chosen from 8 men and 5 women so as to
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07 Apr 2021, 06:15
Whenever you deal with this types of questions, keep the following in mind:
Step 1/ go for a calculation as though there are no any restrictions
Step 2: calculate for the restriction that is required
Final step: Subtract the calculated value you got for the restriction(step 2) from the value you got for Step 1(without restrictions).
The easiest way to solve imo is the following:
At least 2 M and 3 women can be interpreted as the following:
Scenario 1: 2M and 4W, which together makes up the 6 members
8C2*5C4= 140
or(in probability means +)
Scenario 2: 3M and 3W, which together also makes up the 6 members
8C3*5C3= 560
Both scenario's without any restriction: Scenario 1 (140) (or/+) scenario 2 (560)= 700 , which is the same as 8C2*5C4 (or/+) 8C3*5C3 = 700
Usually if we account for the restriction, the result of the that would be subtracted from 700 above, which would make the final answer less than 700, which in this case is choise E.
I tried to understand the way folks above used in the calculation for the restriction, but i couldn't find easy to understand. But as a matter of fact i know that at the end, we have to subtract the result of the restriction from the 700, which again will make the final answer less than 700, Choise E.