SOLUTION: A school organization consists of 5 teachers, 7 parents and 6 students. A subcommittee of this group is formed by choosing 2 teachers, 3 parents and 3 students. How many differen

Algebra ->  Permutations -> SOLUTION: A school organization consists of 5 teachers, 7 parents and 6 students. A subcommittee of this group is formed by choosing 2 teachers, 3 parents and 3 students. How many differen      Log On


   

Question 1106091: A school organization consists of 5 teachers, 7 parents and 6 students. A subcommittee of this group is formed by choosing 2 teachers, 3 parents and 3 students. How many different subcommittees can be formed?
Since there are 5 teachers, I think there are 10 different combinations of 2 teachers for the subcommittee.
Since there are 7 parents, I think there are 35 different combinations of 3 parents for the subcommittee.
Since there are 6 students, I think there are 20 different combinations of students for the subcommittee.
The above work was done writing out all the combinations, which is time consuming. What is the algebraic formula, factorials?
I don't know how to use the above information, that I hope is correct, to figure out the total combinations for the subcommittee without writing out all the combinations, which would be time consuming. What is the algebraic formula? Please show solution.
I saw an answer that showed {5C2}.{7C3}.{6C3}=How many subcommittees can be formed, but I don't know how to solve that problem, unless 210 is the correct answer by multiplying 5.7.6, but I don't think that is correct.
Thank you for any help you can provide to clarify my thoughts.
Many thanks,
Lisa Hart
Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
C%5B5%5D%5E2 = 5%21%2F%282%21%2A3%21%29 = cancel 3! in the numerator and denominator = %285%2A4%29%2F%281%2A2%29 = 20%2F2 = 10.   Was it hard ?


C%5B7%5D%5E3 = 7%21%2F%283%21%2A4%21%29 = cancel 4! in the numerator and denominator = %287%2A6%2A5%29%2F%281%2A2%2A3%29 = 7*5 = 35.    Was it hard ?


C%5B6%5D%5E3 = 6%21%2F%283%21%2A3%21%29 = cancel 3! in the numerator and denominator = %286%2A5%2A4%29%2F%281%2A2%2A3%29 = 5*4 = 20.    Was it hard ?


Now the final answer is  10*35*20.


Can you calculate it on your own ?

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On Combinations and Permutations see the lessons
    - Introduction to Permutations
    - PROOF of the formula on the number of Permutations
    - Problems on Permutations
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on Combinations
    - OVERVIEW of lessons on Permutations and Combinations
in this site.

Also,  you have this free of charge online textbook in ALGEBRA-II in this site
    - ALGEBRA-II - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this online textbook under the topic  "Combinatorics: Combinations and permutations".


Save the link to this textbook together with its description

Free of charge online textbook in ALGEBRA-II
https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson

into your archive and use when it is needed.


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Hello, Lisa,

I got your comment regarding my typo in the lesson "Introduction to permutations".

You are right. It was a typo. I just fixed it.

Thank you for your noticing !



Regarding number of combinations of  "n"  items taken  "m"  at a time,  there is the formula

C%5Bn%5D%5Em = n%21%2F%28m%21%2A%28n-m%29%21%29.

(See my lessons related to Combinations).

You can always cancel one of the factors  m!  or  (n-m)!  in the numerator and denominator.


                    H a p p y   l e a r n i n g  ! !