Using the digits $2$, $3$, $4$, $5$ and $6$, find the sum of all $5$-digit numbers that can be formed such that no two digits are same? Using the digits $2$, $3$, $4$, $5$ and $6$, find the sum of all $5$-digit numbers that can be formed such that no two digits are same?
Why each of the number would be in any place $4!$ times and not $5!$ times?
And how to find the sum? Please can someone explain. There are many questions based on this concept. But I am not able to understand how to find sum of all the $5$-digit numbers.
 A: Here's another way . . .

There are$\;5! = 120\;$such numbers.

For each such number $n$, pair it with what I'll call its complement:$\;88888-n$

Thus, there are$\;5!/2 = 60\;$such pairs.

It follows that the sum is just $60\times 88888 = 5333280$.
A: Fix one digit in any place. Then, using the remaining digits, it is clear that $4!$ and not $5!$ numbers can be formed; there are $5!$ 5-digit numbers to sum up in total.
To get the final sum, consider the sum of the digits in the units place first: $4!(2+3+4+5+6)=24\cdot20=480$. This would be multiplied by $10^n$ for the $10^n$ place, so the sum is $480\cdot11111=5333280$.
A: As an example, how many times does the number $2$ appear in the last position? Or how many numbers can we make of the form $\#\#\#\#2$?
Well we have four positions to fill in with $3$, $4$ ,$5$ and $6$. That's just the same as asking how many four digit numbers are there with those digits? $4!$ is the answer.
A: Here is another way to look at it.  As noted elsewhere, there are 120 different numbers.  Since each place (1s, 10s, 100s, etc.) holds 2, 3, 4, 5, and 6 the exact same number of times, the average for each place is $120 \times 44444 = 5333280$.
