Sanjay 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 is sanjay wants to last digit to be a seven and the order of the digits matters how many ways can the remaining digits be chosen

Easier Explanation:

The answer is 504 because if we are looking for 3 numbers that matter the order, it would be permutation, so, 9 times 8 times 7 is 504.

Advanced Explanation:

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

Let the four digit number be W, X, Y, Z .

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 the repetition 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  = 504.

n= 504.

If you have any questions, feel free to comment below.

Best of Luck to you.

Merry Christmas!

there hope it helps Is (5, 1) a solution to this system of inequalities?

x + 8y < 20

2x + 6y < 16 )=

Step-by-step explanation: