Out of $9$ women and $3$ men, a committee of $6$ members is formed. If at
least $1$ man is always included in committee, then there are __ ways?
Here is the logic I tried:
Case A: $3$ M, $3$ W
Case B: $2$ M, $4$ W
Case C: $1$ M, $5$ W
Using Combinations:
${3\choose1} \times {9\choose5} + {3\choose2} \times {9\choose4} + {3\choose3} \times {9\choose3} = 840$ [the correct answer].
The problem is that I am about to take aptitude exam for college, and I will not be having enough time to solve long calculation, and calculators won't be allowed, too.
So, what is the shortest method or trick for solving this kind of question? I will be having only 1 minute for the question and the test will be MCQ based.
Thanks for your help 🙂
Best Answer
Usually when you are asked a question that contains the words "at least 1" it is quite useful to analyze if it is possible to approach the problem from the "complementary case perspective", for example:
Let A be the number of possible committes, without restrictions. You may work this out, but is as simple as choosing $6$ people from $12$.$$|A|=|\{\text{Number of possible committes}\}|={9+3 \choose 6}={12 \choose 6}=\frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1}=\\=11\cdot7\cdot4\cdot3=924=\text{ posibilites.}$$
Now you want to look at the number of possible committes where there aren't any men. So define $A_0=\{\text{Number of possible committes without men}\}$ which is precisely the complementary subset (in A) of what you are asked.
$$|A_0|={9 \choose 6}=\frac{9\cdot8\cdot7}{3\cdot2\cdot1}=7\cdot4\cdot3=84=\text{ posibilites.}$$
So now you want to figure out the cardinal number of the following set: $\overline{A_{0}}=\{\text{Number of possible committes with at least one man}\}$.
$$|\overline{A_{0}}|=|A|-|A_{0}|=924-84=840\text{ possible committes.}$$