SOLUTION: A club has 18 members, and they need to make a festival committee with 3 people. How many different committees can they make?

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Question 1156923: A club has 18 members, and they need to make a festival committee with 3 people. How many different committees can they make?
Found 2 solutions by ikleyn, josmiceli:
Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.

C%5B18%5D%5E3 = %2818%2A17%2A16%29%2F%281%2A2%2A3%29.

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On Combinations,  see introductory lessons
    - 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.


Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
C( n, r ) = +n%21+%2F+%28+r%21%2A%28+n+-+r+%29%21+%29+
+n+=+18+
+r+=+3+
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I use the formula for combinations because
the order of the 3 people does not matter.
I assume they all have the same ranking
C( 18, 3 ) = +18%21+%2F+%28+3%21%2A%28+18+-+3+%29%21+%29+
C( 18, 3 ) = +18%21+%2F+%28+3%21%2A15%21+%29+
C( 18,3 ) = +%28+18%2A17%2A16+%29+%2F+%28+3%2A2%2A1+%29+
C( 18,3 ) = +3%2A17%2A16+
C( 18,3 ) = +816+
816 different committees possible
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check the math