Use generalized counting principle.

Explanation:

We have 10 digits to choose from {0,1,2,3,4,5,6,7,8,9} for slot A

Then after making that first selection, there are 10-1 = 9 choices left for slot B

Then for slot C, there are 10-2 = 8 choices

Finally for slot D, there are 10-3 = 7 choices

We have a steady countdown: 10,9,8,7

Overall, there are 10*9*8*7 = 5040 different permutations

Order matters because a code like 1234 is different from 4321

Alternative Method:

We can use the nPr permutation formula with n = 10 and r = 4

nPr = (n!)/((n-r)!)

10P4 = (10!)/((10-4)!)

10P4 = (10!)/(6!)

10P4 = (10*9*8*7*6!)/(6!)

10P4 = 10*9*8*7

10P4 = 5040

and we get the same result.

wfirfkfr/jjfi; jn; p/ojkfvlbvmkmgkgkkkgkkg

step-by-step explanation:

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )