mjhjmtu

A company makes mechanical keypad locks.

The keypad is a set of five buttons arranged vertically.

O 
O
O
O
O

The buttons are quite close together. Once a button is pushed in it stays in until the lock is opened or reset.

If the right combination of buttons are pushed then a separate handle can be turned and the door opens. If not the action of turning the handle resets all the buttons so you start again.

The salesman says that there are $545$ different simple 'unlock combinations' which can be 'programmed' into this mechanical lock. Simple unlock combinations are combinations which can be punched into the keypad with a single finger.

Is the salesman correct? why or why not?

Well, let's see. It's possible for 1 to 5 buttons to be pushed to create a combination, so let's calculate the number of possibilities for each individually, assuming you can pick the order of the buttons to be pushed.

1 Button

Well, there are only 5 possible combinations. This isn't hard.

2 Buttons

The number of possible combinations is C(5,2), or 10. You can arrange the order of each possible combination in 2!, or 2 ways. So the number of possible 2-button combinations is 20.

3 Buttons

Since C(5,3) is also 10, there are 10 different combinations. Each of these combinations can be reordered in 3!, or 6 ways. So the number of possible 3-button combinations is 60.

4 Buttons

C(5,4) is 5. So, there are 5 different combinations which each can be rearranged into 4!, or 24 ways. So the number of possible 4-button combinations is 120.

5 Buttons

Obviously, there is only 1 5-digit combination, but that can be rearranged in 5!, or 120 ways.

Total

5+20+60+120+120 = 325 possible combinations. Assuming there isn't any trickery here, the salesman is indeed incorrect.

I believe the salesman may be underestimating the possibilities. In addition to Excited Raichu's 325 arrangements there are the following possibilities:

1)

No buttons pressed = 1 extra possibility

2)

Two neighboring buttons pressed simultaneously. For example (12)345, or 1(23)45, or 12(34)5, or 123(45). For each of these there is one possibility with a single button press (the pair would have to be pressed first). 12 consisting a pair and a single, 24 consisting a pair and 2 singles, and 24 consisting a pair and 3 singles. So an extra 61 combinations for each of the above. = 244 extra possibility

3)

Two pairs of neighboring buttons pressed so either (12)(34)5, or (12)3(45) or 1(23)(45). In each of these cases there is no option in which one button or pair is pressed (as it would be identical to one of the already covered possibilities). 2 options consisting pressing the two pairs (in either order), and 6 options pressing the 2 pairs and the single. So 8 in total for each of the above . = 24 possibilities

4)

If the buttons were really small then there could be 3 or more pressed simultaneously - but I will ignore these calculations as if that were the case it may be difficult to press only one at a time.

So the total could be at least

325 + 1 + 244 + 24 = 594

I see my latest attempt is similar to @excited-raichu's answer, but I have a slightly different count...

There are

32 different combinations after pushing buttons

Of those, there are:

5 combos where 1 button is pushed, 11 combos where 2 buttons are pushed, 9 combos where 3 buttons are pushed, 5 combos where 4 buttons are pushed, and 1 combo where all 5 buttons are pushed.

Each can be pushed in:

#ButtonsPushed factorial ways

So:

5*1 + 11*2 + 9*6 + 5*24 + 1*120 = 321 possible combos (assuming you don't count no buttons being pushed as a combo).