Question

Sonji bought a combination lock that opens with a four-digit number created using the digits 0-9. The same digit cannot be used more than once in the combination.

If sonji wants the last digit to be a 7 and the order of the digits matters, how many ways can the remaining digits be chosen?

A. 84
B. 504
C. 3024
D. 60480

Answer 1

The answer to your question is B just did the test... 

Answer 2

You can use the rule of product from combinatorics to find out how many ways can the remaining digits be chosen.

The remaining digits for the given condition can be chosen in

Option B: 504

What is the rule of product in combinatorics?

If a work A can be done in p ways, and another work B can be done in q ways, then both A and B can be done in  $p \times q$ ways.

Remember that this count doesn't differentiate between order of doing A first or B first then doing other work after the first work.

Thus, doing A then B is considered same as doing B then A

How many lock combinations are possible?

(With given constraints)

Since 7 is already fixed and there is constraint that no two digits can be used, thus, we initially have only 10-1 = 9 digits.

For first place, we can chose 1 digit out of 9 digits in $^9C_1 = 9$ways.

For second place, as one of the digit is used, thus, 1 digit out of 8 digits can be chosen in $^8C_1 = 8$ ways.

For third place, as two of the digits is used, only 7 digits are remaining to chose from. Thus, 1 digit out of 7 remaining digits can be chosen in $^7C_1 = 7$ ways.

Using product rule, we get:

Total possible number of choices for choosing digits for first, second and third position of lock with given constraints is given as $9 \times 8 \times 7 = 504 \: \rm ways$

Thus,

The remaining digits for the given condition can be chosen in

Option B: 504

Learn more about rule of product here:

https://brainly.com/question/2763785