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probabbility

by kksekar » Mon May 07, 2012 8:22 am
A door can be opened only with a security code that consists of five buttons: 1, 2, 3, 4, 5. A code consists of pressing any one button, or any two, or any three, or any four, or all five.
How many possible codes are there? (You are to press all the buttons at once, so the order doesn't matter.) If, to open the door you must press three codes, then how many possible ways are there to open the door? Assume that the same code may be repeated.
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by Bill@VeritasPrep » Mon May 07, 2012 8:29 am
1-digit: 5 codes
2-digit: 5*4/2! = 10 codes
3-digit: 5*4*3/3! = 10 codes
4-digit: 5*4*3*2/4!=5 codes
5-digit: 1 code

For a total of 31 codes.

An easier way to do it is to recognize that we really have an either/or situation for each button: either we press it or we don't. 2 options for each button gives us 2*2*2*2*2=32 possibilities, but that also includes not pressing any buttons (which does not work as a code). 32 - 1 = 31.

If we have to press three codes to open the door and they can be repeated, there are 31^3 possibilities.
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