Question

Sonji bought a combination lock that opens with four-digit number created using the digits 0 through 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?
O 84
• 504
• 3,024
• 60,480

Answer 1

Explanation:

This is what I am thinking. - ,- ,- , - I number goes in each space. There are 10 numbers to choose from and I cannot use a digit more than once. I only have 1 choice for last slot and that is the number 7. I have 9 numbers left . I could put any one of the 9 numbers in the first spot, then there are only 8 numbers to choose from, so I will put 8 in the second spot. I now have only 7 numbers that are left for the third spot.

If I multiple this together 9x8x7x1 = 504