Question

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 60,480 3,024 504

Explanation:

Step1: Determine available digits for first three places

The last digit is fixed as 7, so for the first digit (thousands place), we can't use 7, so we have 9 choices (digits 0 - 9 excluding 7).

Step2: Determine choices for second digit

After choosing the first digit, for the second digit (hundreds place), we can't use the first digit and 7, so we have 8 remaining choices.

Step3: Determine choices for third digit

After choosing the first two digits, for the third digit (tens place), we can't use the first two digits and 7, so we have 7 remaining choices.

Step4: Calculate total number of ways

Using the multiplication principle, the total number of ways to choose the first three digits is \(9\times8\times7=\frac{9!}{(9 - 3)!}=504\)? Wait, no, \(9\times8\times7 = 504\)? Wait, \(9\times8 = 72\), \(72\times7=504\)? Wait, no, wait, the first digit: we can use 0? Wait, the problem says a four - digit number. Wait, if the last digit is 7, the first digit (thousands place) can be 0? Wait, no, a four - digit number cannot start with 0. Wait, I made a mistake. Let's correct it.

Wait, the combination lock is a four - digit number, but does it have to be a four - digit number in the sense of a number (i.e., cannot start with 0) or just a four - digit combination (where 0 can be in the first place)? The problem says "a four - digit number created using the digits 0 through 9" and "the same digit cannot be used more than once". Let's re - evaluate.

If it's a combination (where order matters and 0 can be in the first place), then:

Last digit is fixed as 7. So for the first digit (position 1), we have 9 choices (digits 0 - 9 except 7). For the second digit (position 2), we have 8 choices (digits except 7 and the first digit). For the third digit (position 3), we have 7 choices (digits except 7, first and second digits). Then the number of ways is \(9\times8\times7=\frac{9!}{(9 - 3)!}=9\times8\times7 = 504\)? Wait, no, \(9\times8\times7=504\)? Wait, \(9\times8 = 72\), \(72\times7 = 504\). But wait, if the first digit cannot be 0 (because it's a four - digit number), then for the first digit, we have 8 choices (digits 1 - 9 except 7), then the second digit: 8 choices (digits 0 - 9 except 7 and the first digit), third digit: 7 choices. But that would be \(8\times8\times7\), which is not one of the options. So the problem must consider the combination as a four - digit sequence where 0 can be in the first place (i.e., it's a combination lock, not necessarily a four - digit number in the numerical sense). So going back, first digit: 9 choices (0 - 9 except 7), second: 8 (except 7 and first), third:7 (except 7, first, second). So \(9\times8\times7 = 504\)? Wait, no, \(9\times8\times7=504\)? Wait, \(9\times8 = 72\), \(72\times7 = 504\). Wait, but the options include 504 as one of the options. Wait, but let's check again.

Wait, the total number of permutations of 9 digits (excluding 7) taken 3 at a time (since the last digit is fixed as 7) is \(P(9,3)=\frac{9!}{(9 - 3)!}=\frac{9!}{6!}=9\times8\times7 = 504\). Yes, that's correct. So the number of ways to choose the remaining three digits is 504.

Answer:

The fourth option (the one with 504)