sonji bought a combination lock that opens with a four - digit number created using the digits 0 through 9…

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

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

Answer

Explanation:

Step1: Determine available digits

Since the last - digit is fixed as 7, we have 9 remaining digits (0 - 9 excluding 7) to fill the first three positions.

Step2: Use permutation formula

The number of permutations of (n) distinct objects taken (r) at a time is (P(n,r)=\frac{n!}{(n - r)!}). Here (n = 9) and (r=3). [P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}] [=\frac{9\times8\times7\times6!}{6!}=9\times8\times7 = 504]

Answer:

504