Sonjia bought a combination lock that opens with a 4-digit number created with the numbers 0 through 9. The same digit cannot be used more than once in the combination. If Sonjia wants the last digit to be 7 and the order of the digits matters, how many ways can the remaining digits be chosen?

a

84

b

504

c

60,480

d

3,024

It's b 504

Step-by-step explanation:

Took the test

b. 504

Step-by-step explanation:

In this case, we have to apply the permutation formula, since the order is important and it cannot be repeated. The formula is as follows:

nPr = n! / (n-r) !

From 0 to 9, there are 10 digits, but we must remember that the number 7 has already been used and cannot be repeated, therefore n = 9. And they are 4 digits but only 3 are needed, therefore r = 3.

Replacing:

9P3 = 9! / (9-3) !

nPr = 504

Therefore the answer is b.