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jimthompson5910 (jim_thompson5910):
do you have any ideas on the first step?
OpenStudy (anonymous):
\[_{10} P _{6}\]
OpenStudy (anonymous):
no
jimthompson5910 (jim_thompson5910):
Are you familiar with the permutation formula?
OpenStudy (anonymous):
no
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jimthompson5910 (jim_thompson5910):
If not, the permutation formula is
n P r = (n!)/((n-r)!)
where n! means n! = n*(n-1)*(n-2)*...*3*2*1
For example: 6! = 6*5*4*3*2*1
jimthompson5910 (jim_thompson5910):
So in our case, 10 P 6 is
10 P 6 = (10!)/((10-6)!)
since n = 10 and r = 6
jimthompson5910 (jim_thompson5910):
Can you take it from here or do you need more help?
OpenStudy (anonymous):
yes i need more help please
jimthompson5910 (jim_thompson5910):
(10!)/((10-6)!)
will become
(10!)/(4!)
since 10-6 = 4
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jimthompson5910 (jim_thompson5910):
Then you expand out 10! like so
10*9*8*7*6*5*4!
We stop at 4 because the "4!" terms will cancel
So (10!)/(4!)
turns into
(10*9*8*7*6*5*4!)/(4!)
and the "4!" terms will cancel leaving us with 10*9*8*7*6*5
jimthompson5910 (jim_thompson5910):
Last step is to multiply out 10*9*8*7*6*5 to get your answer