A committee consisting of 3 men and 2 women is to be formed out of 6 men and 4 women. In how many ways this can be done if Mr.X and Mrs.Y are not to be included together?
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- How many committees are there if we disregard the restriction? 2. How many committees contain both Mr X and Mrs Y?
– TonyK Feb 03 '15 at 12:42 -
2Please edit your question to include a description of what you've tried and where you're stuck. This will help people give useful answers at an appropriate level. – Barry Cipra Feb 03 '15 at 12:42
2 Answers
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see it as two subgroups,
- take 3 men out of 6
- take 2 women out of 4
and multiply
Correction (overlooked the bit about Mr.X and Mrs.Y)
Then Mr. X is in half the committees counted this way, so is Mrs. Y so in a quarter of them they are both included, so multiply the previous answer by $ 1- \frac{1}{4} $
GOOD Luck
N. F. Taussig
- 76,571
Willemien
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To determine the number of committees of three men and two women that can be formed from a group consisting of six men and four women subject to the restriction that Mr. X and Mrs. Y do not serve together, subtract the number of committees in which they would serve together from the total number of committees that could be formed.
The total number of committees of three men and two women that could be formed from six men and four women is the number of ways of selecting three of the six men and two of the four women, which is $$\binom{6}{3}\binom{4}{2}$$ The number of committees of three men and two women that could be formed on which both Mr. X and Mrs. Y would serve is the number of ways of selecting two of the other five men and one of the other three women, which is $$\binom{5}{2}\binom{3}{1}$$ Therefore, the number of committees that do not include both Mr. X and Mrs. Y is $$\binom{6}{3}\binom{4}{2} - \binom{5}{2}\binom{3}{1}$$
N. F. Taussig
- 76,571