Seven women and nine men are on the faculty in the mathematics department at a s

FobelloE 2021-10-15 Answered
Seven women and nine men are on the faculty in the mathematics department at a school.
a) How many ways are there to select a committee of five members of the department if at least one woman must be on the committee?
b) How many ways are there to select a committee of five members of the department if at least one woman and at least one man must be on the

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Expert Answer

Bertha Stark
Answered 2021-10-16 Author has 3454 answers

Step 1
Definition
Product rule If one event can occur in m ways AND a second event can occur in n ways, then the number of ways that the two events can occur in sequence is then \(\displaystyle{m}\cdot{n}\)
Definition permutation (order is important):
\(\displaystyle{P}{\left({n},{r}\right)}={\frac{{{n}!}}{{{\left({n}-{r}\right)}!}}}\)
Definition combination (order is not important):
\(C(n,r)=\left(\begin{array}{c}n\\ r\end{array}\right)=\frac{n!}{r!(n-r)!}\)
\(\displaystyle\text{with }\ {n}\ne{n}\cdot{\left({n}-{1}\right)}\cdot\ldots\cdot{2}\cdot{1}\)
Step 2
Solution
The order of the committee members does not matter (because they all receive the same position), thus we then need to use the definition of combination
(a) Committees with any number of men/women
We need to select 5 members from the 9+7=16 people
n=16
r=5 Evaluate the definition of a combination:
\(\displaystyle{C}{\left({16},{5}\right)}={\frac{{{16}!}}{{{5}!{\left({16}-{5}\right)}!}}}={\frac{{{16}!}}{{{5}!{11}!}}}={4368}\)
Committees with only men
We need to select 5 members from the 9 men
n=9
r=5
Evaluate the definition of a combination:
\(\displaystyle{C}{\left({9},{5}\right)}={\frac{{{9}!}}{{{5}!{\left({9}-{5}\right)}!}}}={\frac{{{9}!}}{{{5}!{4}!}}}={126}\)
Committees with at least one woman
We need to select 5 members from the 9 men
4368-126=4242
Step 3
(b)Committees with only women
We need to select 5 members from the 7 women
n=7
r=5
Evaluate the definition of a combination:
\(\displaystyle{C}{\left({7},{5}\right)}={\frac{{{7}!}}{{{5}!{\left({7}-{5}\right)}!}}}={\frac{{{7}!}}{{{5}!{\left({7}-{5}\right)}!}}}={\frac{{{7}!}}{{{5}!{2}!}}}={21}\)
Committees with at least one woman and at least one man
The possible committees with at least one women are the possible committees that do not have only men and do not have only women:
4368-126-21=4221
Result
(a)4242
(b)4221

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