SOLUTION: Formula for (a): n(C)r=n!/r!(n-r)! How many 4-person committees can be formed out of a class of 15? I worked out 15C4=15!/(15-4)!4! that equals 15!(11)!4! and that is as far as I g

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Question 1161854: Formula for (a): n(C)r=n!/r!(n-r)! How many 4-person committees can be formed out of a class of 15? I worked out 15C4=15!/(15-4)!4! that equals 15!(11)!4! and that is as far as I got. This formula comes from "Combinations of n Objects Taken r at a Time."
(b) In how many ways can a 4-person committee be chosen?

Found 3 solutions by solver91311, ikleyn, MathTherapy:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!


, and





You can do the rest of the arithmetic.


John

My calculator said it, I believe it, that settles it


Answer by ikleyn(52818)   (Show Source): You can put this solution on YOUR website!
.

Looking into your post,  I see that your knowledge on the subject is undistinguished from zero.

If you  REALLY  want to learn something on this subject  (or  "start to learn",  saying gently),  you may start
reading the introductory lessons on combinations
    - 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 MathTherapy(10555)   (Show Source): You can put this solution on YOUR website!
Formula for (a): n(C)r=n!/r!(n-r)! How many 4-person committees can be formed out of a class of 15? I worked out 15C4=15!/(15-4)!4! that equals 15!(11)!4! and that is as far as I got. This formula comes from "Combinations of n Objects Taken r at a Time."
(b) In how many ways can a 4-person committee be chosen?
It's NOT 15!(11)!4!, but  instead.
And,
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