Suppose you must choose a 3 digit code for your locker using the digits 0 through 9 and no digit can be used more than once. how many 3-digit codes are possible?

1st digit can be 0-9 = 10 numbers

2nd digit can be 10-1 numbers = 9 numbers

3rd digit can be 10-2 = 8 numbers

10 x 9 x 8 = 720 codes are possible

To solve this problem, you have to use

P

E

M

D

A

S

Which states the order in which you solve the equation.

P (parentheses)

E (exponents)

M (multiplication)

D (division)

A (addition)

S (subtraction)

So first, multiply 3 and 6. Because there are no parentheses or exponents, you skip to multiplying. You should get 18.

Second, you need to divide 6 from 12. You should get 2.

The last thing you need to do is subtract 2 from 18 and you should get 16 for your final answer.