10 buttons, each can be only pressed once.

[helpmeplease37475]

How many different combinations can 10 buttons have, when each button can be only pressed once per combination. Buttons dont have to be pressed. For example, the combination 1574 will work, but 48398 wont.
So far, my only answer has been 403200 different combinations, just want to confirm this here.

 

[Deleted member 4993]

How many different combinations can 10 buttons have, when each button can be only pressed once per combination. Buttons dont have to be pressed. For example, the combination 1574 will work, but 48398 wont.
So far, my only answer has been 403200 different combinations, just want to confirm this here.

Can you please show your work and tell us how you calculated that number?

 

[pka]

How many different combinations can 10 buttons have, when each button can be only pressed once per combination. Buttons dont have to be pressed. For example, the combination 1574 will work, but 48398 wont.
So far, my only answer has been 403200 different combinations, just want to confirm this here.

Are we to assume that at least one button is pressed?
You used the word combinations which has a very clear definition in counting theory, it means the order makes no difference.
That is: \(\displaystyle 1574\text{ is the same as }5714\). Given your suggested answer, I do not think you mean that.
If order does not make any difference the the answer is \(\displaystyle 2^{10}-1\)

If you mean that order does make a difference then \(\displaystyle \sum\limits_{k = 1}^{10} {\frac{{10!}}{{(10 - k)!}}} \) SEE THIS

Please tell us exactly what you meant.

 

[helpmeplease37475]

Are we to assume that at least one button is pressed?
You used the word combinations which has a very clear definition in counting theory, it means the order makes no difference.
That is: \(\displaystyle 1574\text{ is the same as }5714\). Given your suggested answer, I do not think you mean that.
If order does not make any difference the the answer is \(\displaystyle 2^{10}-1\)

If you mean that order does make a difference then \(\displaystyle \sum\limits_{k = 1}^{10} {\frac{{10!}}{{(10 - k)!}}} \) SEE THIS

Please tell us exactly what you meant.

Yeah, my math was way off. I was thinking about the kinda combination where the order doesnt matter, but I worded it wrong and counted it wrong.