How many 3-person committees can be formed in a club with 8 members?
Actually, you can solve this using a scientific calculator and use the combination function. However, if you want to solve it manually, (I think) the formula for combination is (n!) / ((n-m)! X m!), where n denotes the total number of members, while m denotes the number of person committees. (!) is factorial. Factorial examples: 3! = 3 X 2 X 1 = 6 5! = 5 X 4 X 3 X 2 X 1 = 120 For example in your problem: n = 8, and m = 3. therefore: (8!) / ((8-3)! X 3!) = 56 Anyway.. just try the formula. It might help.
The 12 is made up of 2 groups. There is a group of Carmen, and a group of not Carmen. The group of Carmen has 1 person in it, the group of not Carmen has 11 people in it.We want to know how many ways there are to choose 1 person from the Carmen group, and 4 people from the not Carmen group.[math]\dbinom{1}{1} \dbinom{11}{4} = \dfrac{11 \cdot 10 \cdot 9 \cdot 8}{4 \cdot 3 \cdot 2 \cdot 1} = 11 \cdot 5 \cdot 3 \cdot 2 = 330[/math]
You would first already include that particular woman, so then all that’s left is how many committees of 4 can be formed from 8 men and 3 women. But, since there’s no restriction on men and women, the question is how many committees of 4 can be formed from 11 people:[math]==>\binom{11}{4}[/math]=[math]330[/math]so your answer is 330 ways
How many different committees of 3 people can be formed from a pool of 7 people?
7c3 so 35 you're not worried about order unlike the 7*6*5=210 the committee is just formed of 3 people in any order. 123 is the same as 321 and 231 and 132 and 312 and 213 so you must divide 7*6*5* by 3! or 6 since each case has 6 instances.
Number of distinct 4 person committees that can be selected from among 25 people = 25!/(21!)(4!) = 12,650.Edit: If the question is intended to mean the number of ways (permutations) to select 4 person committees from among 25 people then most of the 4 person committees will be composed of the same people but ordered differently and the number of ways would = 25!/21! = 303,600. I’m guessing this was likely not the intent of the question.If the question is asking how many committees of 4 people can be selected from among 25 people with each such committee existing at the same time the answer is obviously just 6 with a single spare person. I’m guessing this was also not the intent of the question.
How many different four-person committees can be formed from a group of six people?
You can rephrase the problem as how many different two-person committees since every four-person committee has a corresponding two-person committee. The first person in the two-person committee can be picked in one of the 6 ways, and the second person can be picked in one of 5 ways (because we will have exclude the first person). That gives a total of 30 possibilities. But if you look carefully, each two-person committee is double counted and hence the answer is 15. If you are aware of combinatorial theory, the answer is 6 choose 4 = 6 choose 2 = 6*5/2 = 15. The idea is that first See http://en.wikipedia.org/wiki/Binomial_coefficient#Combinatorial_interpretation
How many committees can be formed with 5 members having 6 men and 5 women if at least 1 men always included, and 1 particular woman excluded?There are [math]6[/math] men and [math]5-1=4[/math] women because a certain woman is excluded.The five-person committee has to have at least one man. Then this committee can have anywhere from [math]1[/math] man and the [math]4[/math] women up to [math]5[/math] men and none of the women. So, there are[math]\displaystyle \binom{6}{1}\binom{4}{4}+\binom{6}{2}\binom{4}{3}+\binom{6}{3}\binom{4}{2}+\binom{6}{4}\binom{4}{1}+\binom{6}{5}\binom{4}{0}[/math][math]=(6)(1)+(15)(4)+(20)(6)+(15)(4)+(6)(1)[/math][math]=6+60+120+60+6[/math][math]=12+120+120=12+240[/math]=[math]252[/math]possible committees can be formed.