How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be only pick from the rest, so the number of choices only 5
Third digit can be only pick from the rest, so the number of choices only 4
The total number of choices is $6\cdot 4\cdot 5\cdot 4= 480$
So, is my solution true? Or I miss something?
Thanks
 A: For an even no. last digit needs to be even = 4 choices
Total 4 digit even numbers = (Total even numbers) - (Total 3 digit even numbers)
Total even numbers: $4\cdot 6\cdot 5\cdot 4 = 480$
Total 3-digit even numbers: $3\cdot 5\cdot 4\cdot 1 = 60$
Total 4-digit even numbers: $480 - 60 = 420$
A: Consider two cases:
Case 1: first digit even
There are three ways to find the first digit, if the first digit is even.
There are 2,4, and 6.
There are three ways to find the last digit, because one is taken for the first digit.
There are 5 ways to find the second digit and 4 ways to find the third digit.
So, the total ways is: 3*5*4*3= 180 ways
Case 2: first digit odd
There are three ways to choose the first digit, there are: 1,3,5
There are 4 ways to choose the last digit, there are 0,2,4,6
There are 5 ways to choose the second digit and 4 ways to choose the third digit.
So, the total ways is: 3*5*4*4= 240 ways
Total ways: case 1+ case 2= 180+240= 420 ways
A: You are missing that you might have only three choices for the last digit, depending on your choice of the first digit.  You should split into two cases:  first digit even and first digit odd.  Then your approach goes through nicely.  By the way, it is number of choices, not probability.
A: Consider cases:


*

*The last digit is $0$.  Then you have six choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $6 \cdot 5 \cdot 4 \cdot 1 = 120$ four digit even numbers that end in $0$.

*The last digit is $2$, $4$, or $6$.  Since the first digit cannot be zero, you have five choices for the first digit, five choices for the second digit, and four choices for the third digit, giving $5 \cdot 5 \cdot 4 \cdot 3 = 300$ four digit even numbers that end in $2$, $4$, or $6$.  
That exhausts the possibilities, so there are $120 + 300 = 420$ even four digit numbers that can be formed using the digits $0, 1, 2, 3, 4, 5, 6$.
A: Make two cases for this problem
Case 1: fix 0 in last position, so last position can be filled in 1 way. Then first place can be filled in 6 ways, second place can be filled in 5 ways and son on, so total of this case will be 6x5x4x1 = 120
Case 2: Since all even numbers (2,4,6) will come is last position, it can be filled in 3 ways, then first place can be filled by 3 odd numbers + 2 even numbers  (since one even number is fixed in last position), second position can be filled by 4 digits + 0 that is 5 digits, third place can be filled by 4 digits and last place by 3, so total of this case will be 5x5x4x3 = 300
total required numbers = case 1 + case 2 = 120+300 = 420.
A: Without considering different cases.
This is another way to get the same answer as already answerd above.
Here we start by finding the total amount of the four-digit numbers with distinct digits, then finding the amount of odd digits, filling the same criteria, and last, we subtract the odd numbers from the total, to find the even numbers filling the criteria.
We have totally seven digits. We know that the digit zero cannot be placed at the position representing the thousand position. That leaves us with six digits to choose from for this position and that can be made in
$P(6,1) = \frac{6!}{(6-1)!} = \frac{6!}{5!}=6$, different ways.
Now, we have three positions left to fill and six digits to choose from, including the digit zero, which can be placed anywhere in the remaining positions. This choice can be done in
$P(6,3) = \frac{6!}{(6-3)!} = \frac{6!}{3!}=6 \cdot5 \cdot4$, different ways.
Finally, with distinct digits, there is
$6 \cdot6  \cdot5 \cdot4 =720$
four-digit numbers to be constructed.
We know that the amount of odd numbers plus the amount of even numbers equal the total amount of the 720 four-digit numbers.
The odd numbers are 1, 3 and 5. The question to be asked is how many of the 720 are odd?
The digit to fill the unit position can only be chosen from the digits 1, 3 or 5, and this choice can be made in
$P(3,1)=\frac{3!}{(3-1)!}=\frac{3!}{2!}=3$, different ways.
To choose the digit filling the thousand position, we have five valid digits to choose from, since the zero digit is not valid for this position. The choice can be made in
$P(5,1)=\frac{5!}{(5-1)!}=\frac{5!}{4!}=5$, different ways.
Now we are left with two positions to fill, the hundred and tenth position, and five digits to choose from, now including the zero digit. The choice for these two positions can be made in
$P(5,2)=\frac{5!}{(5-2)!}=\frac{5!}{3!}=5 \cdot4=20$, different ways.
The total number of odd four-digit numbers, with distinct digits are
$5 \cdot5  \cdot4 \cdot3 =300$.
Now, we can answer the question how many of these 720, four-digit numbers, are even, by the subtraction
$720-300=420$.
The even numbers are 420.
