The number of three-digit numbers with distinct digits that can be formed using the digits 1, 2, 3, 5, 8, and 9 is.........? . The probability that both the first digit and the last digit of the three-digit number are even numbers is.........? .

Question

anonymous · Answer

The number of combinations of $k$ distinct elements from a set of $n$ elements is given by$$\left(\begin{matrix}n \ k\end{matrix}\right) = \frac{ n! }{ k!\left( n-k \right)!}$$In this problem, $n=6$ and $k=3$

anonymous · Answer

alright thanks...

anonymous · Answer

nah dude i got the rest...  it's all good

anonymous · Answer

moderator?

anonymous · Answer

Your welcome. The probability part is straight forward as there are only two possibilities; either the first digit is 2 and the last one is 8, or vice versa

anonymous · Answer

alright sounds good... you have a goodnight, thanks again @ospreytriple

anonymous · Answer

moderator... answer me

anonymous · Answer

@Kainui

anonymous · Answer

answer me, instead of being creepy watching my post @Kainui