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Thread: Combinatorics problem

  1. #1
    Senior Member Vinod's Avatar
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    Combinatorics problem

    Hello,
    I want to find the probability that among four random digits there occur 3,2,1 or 0 repetitions.

    Solutions. I calculated probability for 0 repetitions= 0.504 and for three repetitions =0.001. But my answer for the probability of two amd one repetitions is incorrect. Author gave the answer 0.063 for the probability of two repetitions and 0.432 for the probability of one repetition. I don't understand how the author calculated that? If anyone knows the correct answer, reply this thread.
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    Re: Combinatorics problem

    Quote Originally Posted by Vinod View Post
    Hello,
    I want to find the probability that among four random digits there occur 3,2,1 or 0 repetitions.

    Solutions. I calculated probability for 0 repetitions= 0.504 and for three repetitions =0.001. But my answer for the probability of two amd one repetitions is incorrect. Author gave the answer 0.063 for the probability of two repetitions and 0.432 for the probability of one repetition. I don't understand how the author calculated that? If anyone knows the correct answer, reply this thread.
    How did you calculate these probabilities? Does order matter? The way you are describing it, order does not matter, but the probabilities you are suggesting indicate that order does matter. Is it ok for the digits to begin with a 0? Assuming order does matter and it is ok for the digits to begin with a 0, here is the probability for exactly one repetition:

    You have 10 digits. Choose 3. You have 3 chosen digits. Choose 1 to be repeated. You have four digits, two are the same. You need to permute them. Here is the probability (again, assuming order matters and the four digits may begin with zero):

    $$\dfrac{ \dbinom{10}{3}\dbinom{3}{1}\dfrac{4!}{1!1!2!} }{10^4}$$

    For two repeated digits, you have two scenarios. One: a single digit is repeated twice. The other, each of two digits is repeated once (for a total of two repetitions). How about you try to figure out the probability of each case?
    Last edited by SlipEternal; Oct 8th 2018 at 06:34 AM.
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  3. #3
    Senior Member Vinod's Avatar
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    Re: Combinatorics problem

    Quote Originally Posted by SlipEternal View Post
    How did you calculate these probabilities? Does order matter? The way you are describing it, order does not matter, but the probabilities you are suggesting indicate that order does matter. Is it ok for the digits to begin with a 0? Assuming order does matter and it is ok for the digits to begin with a 0, here is the probability for exactly one repetition:

    You have 10 digits. Choose 3. You have 3 chosen digits. Choose 1 to be repeated. You have four digits, two are the same. You need to permute them. Here is the probability (again, assuming order matters and the four digits may begin with zero):

    $$\dfrac{ \dbinom{10}{3}\dbinom{3}{1}\dfrac{4!}{1!1!2!} }{10^4}$$

    For two repeated digits, you have two scenarios. One: a single digit is repeated twice. The other, each of two digits is repeated once (for a total of two repetitions). How about you try to figure out the probability of each case?
    Hello,
    No doubt, your answer is correct for one repetition of random digits.I am trying to compute probability of two repetitions of random digits.If I don't get the correct answer,I again approach MHF.
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    Re: Combinatorics problem

    Quote Originally Posted by Vinod View Post
    Hello,
    No doubt, your answer is correct for one repetition of random digits.I am trying to compute probability of two repetitions of random digits.If I don't get the correct answer,I again approach MHF.
    Sounds good. If you have trouble, please post exactly what you did so we can help.
    Last edited by SlipEternal; Oct 8th 2018 at 08:31 AM.
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  5. #5
    Senior Member Vinod's Avatar
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    Re: Combinatorics problem

    Quote Originally Posted by SlipEternal View Post
    Sounds good. If you have trouble, please post exactly what you did so we can help.
    Hello,
    I calculated the answer for probability of two repetitions of random digits manually. A)If we take two digits, they can be written in 0001,0010,0100,1000. So total there are four ways. Now there are nine such pairs such as (0,1),(0,2)...(0,9) and ten random digits (0,1,2...,9).So there are total 4*9*10=360 numbers in which two repetitions of single digit can appear.

    B)Let us take another case in which there are one repetition of one digit and again one repetition of another digit. e.g. take 0 and 1. we can write 0,1 in following 6 different ways. 0011,0101,1001,1010,1100,0110.There are such nine pairs(0,1),(0,2)...(0,9). Hence total such numbers are 9*6=54. For the nest pair 1 and 2,we have 48 numbers (6*8).So total we have 54+48+42+36+30+24+18+12+6=270.

    Hence Total numbers of two repetitions of random digits are A)+ B)=630. Hence it's probability is $\frac{630}{10000}$.
    Now, I cann't express my answer in binomial formulas. Would you express this answer in binomial formulas?
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    Re: Combinatorics problem

    Quote Originally Posted by Vinod View Post
    Hello,
    I calculated the answer for probability of two repetitions of random digits manually. A)If we take two digits, they can be written in 0001,0010,0100,1000. So total there are four ways. Now there are nine such pairs such as (0,1),(0,2)...(0,9) and ten random digits (0,1,2...,9).So there are total 4*9*10=360 numbers in which two repetitions of single digit can appear.

    B)Let us take another case in which there are one repetition of one digit and again one repetition of another digit. e.g. take 0 and 1. we can write 0,1 in following 6 different ways. 0011,0101,1001,1010,1100,0110.There are such nine pairs(0,1),(0,2)...(0,9). Hence total such numbers are 9*6=54. For the nest pair 1 and 2,we have 48 numbers (6*8).So total we have 54+48+42+36+30+24+18+12+6=270.

    Hence Total numbers of two repetitions of random digits are A)+ B)=630. Hence it's probability is $\frac{630}{10000}$.
    Now, I cann't express my answer in binomial formulas. Would you express this answer in binomial formulas?
    For 1 digit repeated twice:
    From 10 digits, choose 2: $\dbinom{10}{2}$.
    From the two chosen digits, choose 1 to be repeated: $\dbinom{2}{1}$.
    Permute the digits: $\dfrac{4!}{1!3!}$.
    Probability:
    $$\dfrac{\dbinom{10}{2}\dbinom{2}{1}\dfrac{4!}{1!3 !}}{10^4}$$

    For two digits, each repeated once:
    Choose two digits from 10: $\dbinom{10}{2}$.
    Permute the digits: $\dfrac{4!}{2!2!}$.
    Probability:
    $$\dfrac{\dbinom{10}{2}\dfrac{4!}{2!2!}}{10^4}$$
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