Evaluate P(3, 0).
Is there an equation?
Im doing Probability and Statistics in math
Like 7! over 8!6!
I have also just started to learn Probability :o
I hate probability it dosent sick in!
No it's nice,I like it.
You mean permutations? P(m, n) = m!/(m - n)!
Questions like this i have just started : ^10P3 and ^7C4. @MKTY
\[\bf P(3,0)=_{3}P _{0}=1\]
It's just the number of permutations where as "C" would imply combinations.
P(n,0) is defined as 1 since NOT picking anything is one option
Ya permutations
but what is P(3, 0), is it one?
3!/3! ....
1 ok
So now what would C(10, 3) be?
P(10,3) ------- 3!
So C(10, 7)?
consider the P to mean "pick" P(n,r) refers to starting at n, and picking r numbers: P(10,3) refers to: 10.9.8 3! can be thought of as P(3,3) if need be soo 10.9.8 ------- = C(10,3) 3.2.1
10.9.8 ------ 3.2.1 10.9.8 ------ 3.2 10.3.8 ------ 2 10.3.4 = 10.12 = 120
i find keeping it all factored helps to notice what parts can cancel or simplify out
ya
C(10,7) = P(10,7)/7! 10.9.8.7.6.5.4 -------------- 7.6.5.4.3.2.1 10.9.8 ------ 3.2.1 so C(10,7) is equal to C(10,3)
Nice, I did it right and your examples made it easyer for me to do it next time thanks
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