Question

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

Answer 1

It’s b 504

Explanation:

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Answer 2

b. 504

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.