Combinations

c(5,1)+c(5,2)+c(5,3)+c(5,4)+c(5,5)=5+10+10+5+1=31

paras747 wrote:a) 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.)??

@paras
We'll consider each scenario for this problem

1. Where the code consists of only 1 digit. We could have 5 possible codes
2. Where the code consists of 2 digits. Since all the digits have to be keyed in at once there is no possibility of repetition. Both the digits have to be unique. So for the first digit we have 5 options and for the second we have 4 options. 5*4=20. However, we have considered the different orders as well which we need to convert back into the unordered combinations by dividing by 2!. 20/2=10 possible codes
3. Where the code consists of 3 digits. Using the principles mentioned in statement 2 we have 5*4*3/3!=10 possible codes
4. Where the code consists of 4 digits. We have 5*4*3*2/4!=5possible codes
5. Where the code consists of 5 digits. We have 5*4*3*2*1/5!=1 possible codes

Summing up we have 5+10+10+5+1=[spoiler]31 possible codes[/spoiler]

Thanks for the explanation. That actually cleared things up.

Wow Ian...That was really cool