If no digit appears more than once, how many 3 digit numbers can be formed from the digits 2,4,5,7,9
So how many different ways can you permute 3 numbers from a set of 5?
does order matter? if not than its a combination
Well it asks for 3 digit numbers, so I think the order must matter because 247 is a different 3 digit number than 274
if 2,4,5 is the same as 2,5,4 then its \[\left(\begin{matrix}5 \\ 3\end{matrix}\right)\]
all it said is if no digit appeared more than once, I'm assuming order doesn't matter since 2,5,4 is the same as 5,2,4 bc each digit appears only once in both cases
without replacement is my take on it
"how many 3 digit numbers can be formed"
the logic goes, for me at least: there are 5 ways to pick the first digit there are 4 ways left to pick the second digit and there are 3 ways left to pick the last digit
It's a permutation without repetition problem.
cool
yeah it's a permutation
Isn't it 5!/2!3!
That's combinations. Permutations is 5!/(5-3)!
no thats a combination
5*4*3= 60
I tend to mix them up, lol
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