Your lock looks a lot like a Kaba Simplex. This lock allows for codes where the order in which the buttons are pressed matters. Also, it allows for buttons needing to be pressed in unison, or not at all. Each button can be pressed only once, though.
Single Buttons Codes
Let's start by just pressing a single button at a time. Each code can be anywhere from 1–5 buttons long.
The number of single button codes is simply $5$ since that's the number of buttons.
For the first button, you can pick any of the 5 buttons, which leaves you 4 for the second button. This allows for a total of $5 \times 4 = 20$ two-button combinations. This is the number of permutations of 2 out of 5, or $_5P_2$
$_5P_3 = 5 \times 4 \times 3 = 60$
$_5P_4 = 5 \times 4 \times 3 \times 2 = 120$
$_5P_5 = 5 \times 4 \times 3 \times 2 \times 1 = 5! = 120$
The total number of codes you can enter this way is $5 + 20 + 60 + 120 + 120 = 325$ which is a lot less than the $545$ promised.
Codes with Pairs
The buttons are quite close together.
so we can push two adjacent buttons together. After all, the Simplex allows for buttons to be pressed simultaneously with one finger. This adds a number of multi-button codes.
This adds $4$ codes, since there are 4 pairs of adjacent buttons that can be pressed together with one finger.
There are again $4$ possible pairs, which leave $3$ options for the single button. All of these can be used in two ways: either the pair first, or the single button first.
That gives us a total of $4 \times 3 \times 2 = 24$ codes with a pair and a single button.
Here we have two options. We can either use two pairs, or a pair and two single buttons.
With two pairs, we have $3$ options for the button we leave unpushed: either an outer button, or the middle button. We can push the top pair first, or the bottom pair first, so we have a total of $3 \times 2 = 6$ codes with two pairs.
With a single pair and two single buttons, we have $4$ options for the pair, then $3$ for the unpushed button, so that's $4 \times 3 = 12$ combinations. We can push the pair first, last, or in between the single buttons, and we can either push the top single button first or last. So that's $12 \times 3 \times 2 = 72$ codes.
That gives us a total of $6 + 72 = 78$ codes for 4 buttons with pairs.
Again, we have two options: two pairs and a single, or one pair and three singles.
With two pairs, the single button can be in $3$ places (just like the unpushed button). We can push the single button first, last, or in between the pairs, and we can push the top pair first or last. This gives us $3 \times 3 \times 2 = 18$ codes.
With a single pair, that pair can be in $4$ positions, the rest of the buttons being the single ones. We can push those single buttons in $3 \times 2 \times 1 = 6$ different orders, and we can push the pair first, second, third, or last. This gives us $4 \times 6 \times 4 = 96$ codes.
The total for 5 buttons with pairs comes to $18 + 96 = 124$ different codes.
Adding these all up gives us $4 + 24 + 78 + 124 = 230$ different codes when adding pairs into the mix.
Adding this to the number of single button codes, we get $325 + 230 = 545$ codes as promised.
Codes with Triplets
But can we do more? If we're really fat-fingered, or if the buttons are really close together, maybe we can try pressing 3 adjacent buttons at once.
There are just $3$ triplets we can pick here.
We have the same $3$ triplets and $2$ different single buttons to pick, and we can push the single button either first or last, for a total of $3 \times 2 \times 2 = 12$ different codes.
We can combine a triplet with either a pair, or two single buttons.
There are $2$ different ways to divide the 5 buttons into a pair and a triplet, and $2$ orders in which to press these, so that's $2 \times 2 = 4$ codes.
Or we can pick our triplet in $3$ ways, our remaining single buttons in $2$ different orders, and our triplet either first, last, or in between, so that's $3 \times 2 \times 3 = 18$ codes.
That's $4 + 18 = 22$ different codes.
So that's $3 + 12 + 22 = 37$ additional codes using triplets. If we add this to the $545$ codes we already had, we can even get $545 + 37 = 582$ codes in total.
Codes with Quads
We might as well continue. Let's try pressing 4 buttons at once.
We have the choice of $2$ different quads, and the choice of pressing the remaining button before that, after that, or not at all. That's $2 \times 3 = 6$ codes with quads.
Adding these to what we already had, we get $582 + 6 = 588$ codes in total.
Codes with quints
Let's go all the way and add the last possibility of pressing all 5 buttons at once. There's only $1$ way to do that.
That gives us a final total of $588 + 1 = 589$ possible codes that can theoretically entered using just a single finger.