Question

there is a lock ..only 4 digits are required to open it.if the numbers are (0-9) you can use the numbers more then once...In how many ways are there to open the lock? PLEASE HELP MEDAL!!!

Answer 1

0000 to 9999

Answer 2

9999?

Answer 3

+1

Answer 4

include 0000

Answer 5

yeah the numbers are 0 to 9 and they can repeat

Answer 6

Is this a permutation or combination not sure

Answer 7

http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 8

there are 10 numbers to choose from (0,1,2...,9)

Answer 9

thanks for that site I will definantly check it out...............yeah there are ten number

Answer 10

then what the next step would be?

Answer 11

The problem is a permutations with repetition...
10^4 = 10 000 ways to open

Answer 12

look at the permutations with repetition , it is towards the top of that page

Answer 13

the number of choices for each place, does not reduce if a number is already used, so you have 10 choices for the first space, 10 for the second... and so on

Answer 14

ok I got if....but now what if the number cannot repeat..??

Answer 15

I just need to multiply 9x8x7x6??

Answer 16

then if a number is used , the next space has 1 less to choose from

Answer 17

that is where the factorial thing comes in

Answer 18

but, you want to cut it off depending on how many numbers you are choosing , here is 4

Answer 19

so take 10 factorial, then divide that by (10-4) factorial

Answer 20

that will give you
\[\frac{ 10*9*8*7*6*5*4*3*2*1 }{ 6*5*4*3*2*1 } = 10*9*8*7\]

Answer 21

5040

Answer 22

yeah, if you cant have double numbers in the combination

Answer 23

alright.thanks a lot for you help DanJs

Answer 24

welcome, use that page i linked, it is all there pretty clear.