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all right lets look at number 53 from section 11.2 and id strongly recommend you also watch the video for number 51 in this section because essentially theyre the same question one uses letters and one uses numbers but the procedure is exactly the same when we were dealing with a word and we needed to figure out the number of distinct permutations what we did is we counted all of the letters in the word first so in this number there are one two three four five six seven digits in this number so were going to put a seven factorial in the numerator so however many digits there are its that many factorial that go into the numerator but now we need to account for any numbers that show up more than once so we go through and we look five it only shows up once we dont have to worry about that 4 it shows up twice so we need to put a 2 factorial in the denominator to account for the fact that we have a double number and now we have 1 2 three four sixes so we need to put a four factorial
