Evaluate the permutation. 6 P 2
use the formula for permutations http://www.mathwithlarry.com/lessons/lessonimages/img61.gif
\[\frac{6!}{(6-2)!}\]
So \(\frac{6!}{4!}\)
\(6\times 5=?\)
^What she said.
why is it 6*5?
isnt the answer 720/24?
6 P 2 = 6x5 just expand the factorial by 2 terms
example : 10 P 3 = 10x9x8
Yes, the answer can be 720/24 (because it brings you to the same conclusion)
\[\huge \frac{6*5\cancel{4}*\cancel{3}*\cancel{2}*\cancel{1}}{\cancel{4}*\cancel{3}*\cancel{2}*\cancel{1}}\]
okay thanks yall. can someone help me with this one? Evaluate. (n - 1)!, where n = 4
start by plugging in n = 4
is it 24-1?
If you understood factorial functions, you would understand better.. No, it's not.
nope, remember PEMDAS ? always execute stuff inside parenthesis first
(n-1)! (4-1)! Order of operations dictates you do parentheses first
okay so (4-1)! is 3!
3 and an exclamation is it !!!
yay thank yall!
any day m'lady
i think you want to evaluate the factorial also : 3! = 3x2x1
Do you understand how to do this, Madison?
im pretty sure i do @DangerousJesse . if i was supposed to evaluate 5! it would be 5*4*3*2*1 correct?
Yes! Once you get the concept, it becomes much easier.. Nice job :)
yes exactly ! having known the factorial notation, you just follow ur nose and evaluate everything : 100! = 100*99*89*....*3*2*1
what is the value of \(\huge \frac{(n+1)!}{n!}\)
stop once u reach 1
so when a question is 6 C 4 , how do i determine that?
its a different formula, but similar one. look up ur notes for `combination formula`
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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