1. ## Counting-Permutations

I have a question asking me to find:
The sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions

I realise that the digits can be chosen in (4)_4=4!=24 ways since repetition is not permitted and order does matter

Any help would be great

2. Let the 24 numbers be $a_1,\dots,a_{24}$. Further, let the $i$th number $a_i$ be $1000b_{i1}+100b_{i2}+10b_{i3}+b_{i4}$ where all $b_{ij}$ are in {2,4,6,8}. Let's find $S_1=\sum_{i=1}^{24}b_{i1}$. Each of the four numbers is encountered 6 times, so the $S_1 = 6(2 + 4 + 6 + 8) = 120$. Therefore, the first digits of all numbers contribute 120 * 1000 to the whole sum. Similarly, the second digits contribute 120 * 100 and so on.

3. Originally Posted by qwerty10
I have a question asking me to find:
The sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions
Think about it, each of those four numbers will be in each decimal place six times. The ones column adds up to $6(2+4+6+8)=120.$
Check out $\displaystyle120\left( {\sum\limits_{k = 0}^3 {10^k } } \right) = ~?$.

4. Hello, qwerty10!

Find the sum of all 4 digit numbers containing the digits 2,4,6,8 without repetitions.

List the $4! = 24$ permutations of the four digits
,. . and consider their sum.

. . $\begin{array}{cc}
&2468 \\ &2486 \\ &2648 \\ &2684 \\ &2846 \\ &2864 \\ &\vdots \\
&8246 \\ &8264 \\ &8426 \\ &8462 \\ &8624 \\ + & 8642 \\ \hline
\end{array}$

We find that each column has: six 2's, six 4's, six 6's, six 8's.

The total of each column is: . $6\!\cdot\!2 + 6\!\cdot\!4 + 6\!\cdot\!6 + 6\!\cdot\!8 \:=\:120$

Hence, the addition has the form:

. . $\begin{array}{cccccc}
&&& 1 & 2 & 0 \\ && 1 & 2 & 0 \\ & 1 & 2 & 0 \\ 1 & 2 & 0 \\ \hline 1 & 3 & 3 & 3 & 2 & 0 \end{array}$

Therefore, the sum is: . $133,\!320$

Edit: Plato beat me to it . . . *sigh*
.