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Thread: binomial problem!

  1. #1
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    binomial problem!

    5 digits numbers is formed by using the digits 0,1,2,3,4 and 5 without repetition. find the probability that the number is divisible by 6?

    pls help out here, confuse!
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  2. #2
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    Re: binomial problem!

    Hello, lawochekel!

    I'll explain it using baby-steps.


    Five-digit numbers are formed by using the digits {0,1,2,3,4,5} without repetition.
    Find the probability that the number is divisible by 6.

    There are $\displaystyle 6!= 720$ possible five-digit numbers.

    To be divisible by 6, the number must be:
    . . (1) divisible by 3 ... the sum of its digits must be divisible by 3.
    . . (2) even ... the last digit must be even.


    There are two sets of digits whose sums are divisible by 3:
    . . $\displaystyle \{0,1,2,4,5\}\:\text{ and }\,\{1,2,3,4,5\}$

    Consider $\displaystyle \{0,1,2,4,5\}$
    The last digit must be even: $\displaystyle \{0,2,4\}$, 3 choices.
    The other four digits can be arranged in $\displaystyle 4!$ ways.
    Hence, there are:.$\displaystyle 3\cdot4! \,=\,72$ numbers divisible by 6.

    Consider $\displaystyle \{1,2,3,4,5\}$
    The last digit must be even: $\displaystyle \{2,4\}$, 2 choices.
    The other four digits can be arranged in $\displaystyle 4!$ ways.
    Hence, there are:.$\displaystyle 2\cdot4! \,=\,48$ more numbers divisible by 6.

    Hence, there are:.$\displaystyle 72 + 48 \,=\,120$ numbers which are divisible by 6.


    Therefore:.$\displaystyle P(\text{div by 6}) \:=\:\frac{120}{720} \:=\:\frac{1}{6}$

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  3. #3
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    Re: binomial problem!

    Quote Originally Posted by Soroban View Post
    There are $\displaystyle 6!= 720$ possible five-digit numbers.
    Are you counting $\displaystyle 05421$ as a five-digit number?
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