Sonji bought a combination lock that opens with a 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? 84 504 3,024 60,480

504 ways.

Step-by-step explanation:

total Combinations from 0-9 digits: {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

let the four digit number be WXYZ

W has 9 combinations because 7 is the last digit we omit it,then 9 combinations are left

X has 8 combinations because 7 is the last digit  and repition of numbers is not allowed,so we remove 7 and W.

Y has 7 combinations because 7,W and X are omitted.

Z has 1 combination because last digit must be 7.

So,total number of ways (n), can be calculated as:

n= 9*8*7*1

n= 504 is answer.

step-by-step explanation:

try remembering the steps

step-by-step explanation:

de-gf

hj=3x+20

ji=15x-64