SOLUTION: You just got a free ticket for a boat ride, and you can bring along 4 friends! Unfortunately, you have 6 friends who want to come along. How many different groups of friends could

Question 1158898: You just got a free ticket for a boat ride, and you can bring along 4 friends! Unfortunately, you have 6 friends who want to come along.
How many different groups of friends could you take with you?
Found 2 solutions by solver91311, ikleyn:
Answer by solver91311(24713)   (Show Source): You can put this solution on YOUR website!

If you have 6 friends, there are 6 ways to choose the first person who comes with you. For each of those 6, there are 5 ways to choose the second person. Hence there are 6 X 5 = 30 ways to choose the first two. Then, for each of those 30 ways to choose the first two, there are 4 ways to choose the third person...and so on.

John

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

Answer by ikleyn(52788)   (Show Source): You can put this solution on YOUR website!
.

The order of friends inside the selected group DOES NOT MATTER. Therefore, the answer is the number of COMBINATIONS = = 3*5 = 15. ANSWER Not the number of permutations, as the other tutor mistakenly thinks.

-------------------

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.

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