# Math Help - Counting-Permutations

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

The final answer is 133320

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
The final answer is 133320
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.

The answer is: 133,320.

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*
.