how many 3-digit numbers can be formed using only the digits 1 to 7, if the number 2 must be included? (Repetitions are allowed)

The books says 127. Could someone please explain this? Thanks

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asked by Jim
  1. Since repetition is allowed, and the 2 must be included, it is easy to forget that the 2 can show up once, twice or even three times

    case 1: only one 2
    number of ways = 3(6)(6) = 108
    (2XX, X2X, XX2, where X is one of the other 6 digits)

    case 2: two 2's
    number of ways = 3(6) = 18
    (22X, 2X2, X22, where X is one of the other 6 digits)

    case 3: three 2's , namely 222
    number of ways = 1

    total = 108+18+1 = 127

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    posted by Reiny
  2. 127=7^3-6^3

    7^3 all 3-digit numbers, formed using
    6^3 ... using 1,3,4,5,6,7

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    posted by Mgraph
  3. Mgraph
    Very nice, good logic.

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    posted by Reiny

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