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Mathematics 16 Online
OpenStudy (anonymous):

can anyone help me with Combinations? ive never seen anything like this. 9C1

OpenStudy (lgbasallote):

\[9C1 = \frac{9!}{1!(9-1)!}\] ever seen those before?

OpenStudy (anonymous):

never. whats the exclamation point mean?

jimthompson5910 (jim_thompson5910):

Something like 6! means 6! = 6*5*4*3*2*1 You take the number in front of the ! and you multiply it by every integer less than it until you get to 1

jimthompson5910 (jim_thompson5910):

So 9! = 9*8*7*6*5*4*3*2*1

jimthompson5910 (jim_thompson5910):

and it's read as "9 factorial"

OpenStudy (anonymous):

wow that seems like a long process. but alright so then with the (9-1)! i would do 8!?

jimthompson5910 (jim_thompson5910):

yes

jimthompson5910 (jim_thompson5910):

you evaluate what's inside the parenthesis first, then find the factorial of that

jimthompson5910 (jim_thompson5910):

your calculator should have a ! button to save you time

OpenStudy (anonymous):

ok, so then i guess the denominator simplifies to 1!8! . what do i do from there?

OpenStudy (anonymous):

i dont see the ! button

jimthompson5910 (jim_thompson5910):

1! = 1, so that multiplies with 8! to get 1*8! = 8!

jimthompson5910 (jim_thompson5910):

So 1!8! turns into just 8!

jimthompson5910 (jim_thompson5910):

8! = 8*7*6*5*4*3*2*1 what kind of calculator do you have?

OpenStudy (anonymous):

ohhhh that makes sense. so then it would be 9!/8! and just do the factorial thing and divide? i have a texas instrument TI-30XIIS i think

jimthompson5910 (jim_thompson5910):

yes, find the factorial of each, then divide and simplify

OpenStudy (anonymous):

thats so much easier than i thought it would be. alright, thank you very much!

jimthompson5910 (jim_thompson5910):

btw, I think if you checked the "prb" button, it might be in there

jimthompson5910 (jim_thompson5910):

that's where all the probability functions are

OpenStudy (anonymous):

ohh wow i just checked, its there. wonderful(:

jimthompson5910 (jim_thompson5910):

the factorial, permutation and combinations should all be there

jimthompson5910 (jim_thompson5910):

alright great

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