SOLUTION: for a bowling team of 4, how many different teams are possible from a group of 10?

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Question 1193835: for a bowling team of 4, how many different teams are possible from a group of 10?
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52805) About Me  (Show Source):
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It is  C%5B10%5D%5E4 = %2810%2A9%2A8%2A7%29%2F%281%2A2%2A3%2A4%29 = 210, the number of COMBINATIONS of 10 items/persons, taken 4 at a time.

Solved and explained.

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On  Combinations,  see introductory lessons
    - Introduction to Combinations
    - PROOF of the formula on the number of Combinations
    - Problems on 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 math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Assuming that order doesn't matter, we have 10 C 4 = 210 ways to form the team.
I used the nCr formula as shown in the scratch work below.

n C r = (n!)/(r!(n-r)!)
10 C 4 = (10!)/(4!*(10-4)!)
10 C 4 = (10!)/(4!*6!)
10 C 4 = (10*9*8*7*6!)/(4!*6!)
10 C 4 = (10*9*8*7)/(4!)
10 C 4 = (10*9*8*7)/(4*3*2*1)
10 C 4 = (5040)/(24)
10 C 4 = 210

Notice on the third to last step, I have 10*9*8*7 over top 4*3*2*1
The numerator represents the number of permutations (where order matters) when choosing four people from a pool of ten.
We divide over 4! = 4*3*2*1 = 24 because this is the number of ways to arrange any group of four people. This leads us to the correct count of 210. This is to avoid overcounting.


Answer: 210