How do i answer this permutation question?

[Biochemgirl2002]

Question:
A home security device with 10 buttons is disarmed when three different buttons are pushed in the proper sequence. (No button can be pushed twice.) If the correct code is forgotten, what is the probability of disarming this device?

My attempt:

10!/(10-3)! =( 10x9x8x7x6x5x4x3x2x1)/(7x6x5x4x3x2x1)
= 720

therefore there is a 1/720 chance of getting the right combination.

is this correct?

[Moderator's note: Moved from a technical forum and thus no template.]

 

[jedishrfu]

Yes, it looks correct:

you have 10 choices on the first button, 9 choices on the second and 8 choices on the third:

10*9*8 = 720

Hence you have 1/720 or 0.14% chance of getting it right the first time.