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Combination question

  1. May 7, 2006 #1

    There's a combination exercise that has been bewildering me for some time now: how many divisors does the number 378 have?

    I know it can be done like this: 378=2x3x3x3x7.
    Divisors are as follows: 1,2,3,6,7,9,12,14,18,21,42,54,63,126,189,378. 16 all together.

    But, in essence, it has to be a combinations task, the only catch is that the element 3 is recurrent.

    Is there a formula that takes this aspect into account?

  2. jcsd
  3. May 7, 2006 #2
    Lets say you have three "pools" amongst which you can draw your factors. Let one contain 2^0 and 2^1; one 3^0, 3^1, 3^2, and 3^3; and the last 7^0 and 7^1. Each distinct divisor is made by choosing one number from each pool. So it can easily be seen that the number of total divisors is 2 * 4 * 2 = 16.
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