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11 May, 12:42

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

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  1. 11 May, 14:07
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    504 ways.

    Step-by-step explanation:

    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!
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