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2 problems in combinatorics

  1. Sep 24, 2014 #1
    Hi, there are two problems I can't really solve yet they don't seem that difficult. The two of them seem pretty related to me, I think there's something I'm not getting I'll detail my attempts at solving but any help especially with the steps to the solution would be really appreciated as I have the answers, these are not homework or anything like that just me practicing.

    The first one is: Consider 4 digits, what are the probabilities of getting 1 different digit, 2 of them, 3 of them or 4.
    I have no problem with 1 digit, it is 10/1000. The third I can do also and the 4th, it is especially with two different digits that I struggle.

    So I thought the two digits which can repeat twice can be 0,1;0,2;0,3;0,4;0,5,etc...1,2;1,3;1,4;1,5..etc,2,3;2,4;2,5...etc,2,9; until 8,9 so a total of 45 possibilities.. but out of 10000? because 10^4? Anyways the answer is supposed to be 630.... I don't know how that's possible.. Thanks

    2nd problem

    From the population of five symbols a, b, c, d, e, a sample of size 25 is taken. Find the probability that the sample will contain five symbols of each kind.

    The only thing I could think of was (4/5)^5 x (3/5)^5 x (2/5)^5 x (1/5)^5 which is obviously wrong. Apparently the answer is 25!/(5!^5 X 5^25 ) so if anyone could explain the logic behind this answer would be great! Thank you guys! :)
     
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  3. Sep 24, 2014 #2

    HallsofIvy

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    What, exactly, do you mean by "two different digits"? Different from what? If you number were "1101" would that count as "one different digit" (since the single digit "0" is different from the other digits) or would it count as "two different digits" (since it has "1" and "0" as digits)?
     
  4. Sep 24, 2014 #3
    I had interpreted it as counting as 2 different digits! Or else 4 different would be impossible.
     
  5. Sep 24, 2014 #4

    mathman

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    For your second question, there is an unstated assumption that all the letters are equally probable.
    1/5^25 is the probability for one particular sequence of 25 letters.
    25!/ 5!^5 is the number of sequences with 5 of each letter.
     
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