How many 6-digit numbers can be formed using the digits 1,2,3,4,5,6 and 7, so that the digits do not repeat and the terminal digits should be even? We know that the last digit should be even --> 2,4 or 6
Also, a digit should only be used only once --> 7.6.5.4.3.2 (let us call it 'var1')
So, shouldn't the answer be (var1)x3 ? ( 3  for each number ending with 2, 4 and 6 respectively)
Can someone point out any mistake(s) I may have made ?
Thank you.
 A: The $7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2$ would be the correct answer if we didn't have the final digit being even constraint. But as we do, it makes sense to think of the number in two separate parts: the last digit, and then the first five.
As you say, there are three choices for the terminal digit. But then, for the first five digits, we don't have a choice of seven digits any more - we have the choice of six ($\{1,3,5,7 \}$ or one of the unused two even digits). So the number of options for the first five is $6\cdot 5\cdot 4\cdot 3\cdot 2 = 720$.
The final answer is therefore $3 \cdot 720 = 2160$ possible numbers.
EDIT: Just for fun, I confirmed the answer in Python, so the question can also be answered programmatically, though the above method is much preferable.
count = 0
for i in range(100000, 999999 + 1):
    temp = list(int(x) for x in str(i))
    if len(temp) == len(set(temp)) and temp[-1] % 2 == 0:
        if "0" not in str(i) and "8" not in str(i) and "9" not in str(i):
            count += 1
print("count =", count)

Output: count = 2160
A: Thanks to the help from the comment, if you still fancy reading the answer below, you will get the same answer.
If they used $2$ even numbers from $3$ even numbers in the first $5$ digits and order them, after then they pick up $3$ numbers from the $4$ odd numbers and put them into the remain $3$ positions and need to order them.
$T_1 = (3C2 * 5C2 * 2P1) *(4C3 * 3P3$): 1440
If they choose $1$ even numbers from $3$ even numbers and put them into the first $5$ digits, and they pick up $4$  numbers from $4$ odd numbers and put them into remaining $4$ positions. And they pick up $1$ even number from the $2$ even numbers. 
$T_2$ =  ($3C1 * 5C1) * (4C4 * 4P4) * 2C1$ ):720
$T = T_1 + T_2$.
