Help in the probability....
How many 3-digits numbers can be formed from the digits 0,1,2,3,4,5 and 6 if each digit can be used only once?
And how may of these are even numbers?
Thanks for the help...
Regards!
The number of 3-digits number (including the numbers that start with 0) is $\displaystyle A_6^3=6\cdot 5\cdot 4=120$
But we have to eliminate the numbers that star with 0, wich means $\displaystyle A_5^2=5\cdot 4=20$
So, we have $\displaystyle 120-20=100$ 3-digits numbers.
To count the even numbers we put an even digit on the unit place. There are 4 possibilities: 0, 2, 4, 6.
If the unit digit is 0, there are $\displaystyle A_5^2=20$ even numbers.
If the unit digit is 2, 4 or 6, there are $\displaystyle A_5^2-A_4^1=16$ even numbers.
So, there are $\displaystyle 20+3\cdot 16=20+48=68$ even numbers.