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Mathematics 8 Online
OpenStudy (ellamoyseyuk):

A basketball team has 12 players and the coach wants to pick which 5 players will start today's game. How many combinations are there for the coach to choose from? the answer choice: A. 25 B. 867 C. 792 D. 1,248

OpenStudy (ellamoyseyuk):

@jim_thompson5910 can u help me?

OpenStudy (agent0smith):

This is just a matter of plugging numbers in. Where are you stuck?

OpenStudy (agent0smith):

"12 players and the coach wants to pick which 5 players"

OpenStudy (ellamoyseyuk):

what is the formula for this

OpenStudy (danjs):

when he chooses a person, there are a certain number left in the group to choose, first 25, then 24 after one is chosen, then 23 to choose from.....

OpenStudy (danjs):

sorry there are 12 people to start not 25 for 5 choices then it is.. 12*11*10*9*8

OpenStudy (bunny12):

You can also use the formula to figure this out, but you do need to know what the variables stand for.

OpenStudy (agent0smith):

"12 players and the coach wants to pick which 5 players" From 12 choose 5 You should have the combinations formula.

OpenStudy (agent0smith):

@DanJS it's combinations not permutation.

OpenStudy (ellamoyseyuk):

12/5

OpenStudy (bunny12):

\[nCr=(n!)/((n-r)! * r!)\]

OpenStudy (danjs):

ah right, so there are some teams that are just the same 5 but in a different order so you have to reduce that number of permutations down by how many ways you can arrange the group of 5,

OpenStudy (ellamoyseyuk):

=(12!)/(12-5)!*5

OpenStudy (danjs):

5 people can be arranged 5*4*3*2*1 ways or 5! so you take how many permutations of that group can come out of the 12, and divide it by the doubles(5*4*3*2*1)

OpenStudy (bunny12):

Yes, that's correct

OpenStudy (ellamoyseyuk):

=475200

OpenStudy (bunny12):

Heads up, don't forget the factorial on your 5

OpenStudy (ellamoyseyuk):

the what

OpenStudy (danjs):

\[\frac{ 12*11*10*9*8 }{ 5*4*3*2*1 }\]

OpenStudy (bunny12):

exclamation point on the 5. You forgot it, just wanted to make sure you remembered it, just in case

OpenStudy (ellamoyseyuk):

so the answer is c)

OpenStudy (danjs):

right, 792 ways to pick combination of 5 people out of 12 people

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