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Mathematics 12 Online
OpenStudy (anonymous):

Evaluate the permutation. 6 P 2

OpenStudy (agent0smith):

use the formula for permutations http://www.mathwithlarry.com/lessons/lessonimages/img61.gif

OpenStudy (dangerousjesse):

\[\frac{6!}{(6-2)!}\]

OpenStudy (dangerousjesse):

So \(\frac{6!}{4!}\)

OpenStudy (dangerousjesse):

\(6\times 5=?\)

OpenStudy (anonymous):

^What she said.

OpenStudy (bibby):

why is it 6*5?

OpenStudy (anonymous):

isnt the answer 720/24?

ganeshie8 (ganeshie8):

6 P 2 = 6x5 just expand the factorial by 2 terms

ganeshie8 (ganeshie8):

example : 10 P 3 = 10x9x8

OpenStudy (dangerousjesse):

Yes, the answer can be 720/24 (because it brings you to the same conclusion)

OpenStudy (bibby):

\[\huge \frac{6*5\cancel{4}*\cancel{3}*\cancel{2}*\cancel{1}}{\cancel{4}*\cancel{3}*\cancel{2}*\cancel{1}}\]

OpenStudy (anonymous):

okay thanks yall. can someone help me with this one? Evaluate. (n - 1)!, where n = 4

ganeshie8 (ganeshie8):

start by plugging in n = 4

OpenStudy (anonymous):

is it 24-1?

OpenStudy (dangerousjesse):

If you understood factorial functions, you would understand better.. No, it's not.

ganeshie8 (ganeshie8):

nope, remember PEMDAS ? always execute stuff inside parenthesis first

OpenStudy (bibby):

(n-1)! (4-1)! Order of operations dictates you do parentheses first

OpenStudy (anonymous):

okay so (4-1)! is 3!

ganeshie8 (ganeshie8):

3 and an exclamation is it !!!

OpenStudy (anonymous):

yay thank yall!

OpenStudy (bibby):

any day m'lady

ganeshie8 (ganeshie8):

i think you want to evaluate the factorial also : 3! = 3x2x1

OpenStudy (dangerousjesse):

Do you understand how to do this, Madison?

OpenStudy (anonymous):

im pretty sure i do @DangerousJesse . if i was supposed to evaluate 5! it would be 5*4*3*2*1 correct?

OpenStudy (dangerousjesse):

Yes! Once you get the concept, it becomes much easier.. Nice job :)

ganeshie8 (ganeshie8):

yes exactly ! having known the factorial notation, you just follow ur nose and evaluate everything : 100! = 100*99*89*....*3*2*1

OpenStudy (bibby):

what is the value of \(\huge \frac{(n+1)!}{n!}\)

ganeshie8 (ganeshie8):

stop once u reach 1

OpenStudy (anonymous):

so when a question is 6 C 4 , how do i determine that?

ganeshie8 (ganeshie8):

its a different formula, but similar one. look up ur notes for `combination formula`

OpenStudy (bibby):

meh, I'll just finish up that last thought I had written. \(\huge \frac{(n+1)!}{n!}=\frac{(n+1)*\cancel{n*...*2*1}}{\cancel{n*...*2*1}} \)

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