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Math Help - Permutation (URGENT)

  1. #1
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    Permutation (URGENT)

    Hello all. I seriously need an urgent help from y'all.If you would so kind , could anyone of you please solve these two questions , which is of permutation. My brain couldn't go further :

    "How many numbers greater than 5000 can be formed using he digits 3 , 4 , 5 , 6 and 7 without repetition of digits? How many of these are even numbers?"


    AND

    " How many numbers between 2000 and 8999 (both inclusive) do not have four different digits ? "

    please save me, im almost drowned of solving this kind of things
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  2. #2
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    Hello, discovery!

    Here's the second one . . .


    How many numbers between 2000 and 8999 (both inclusive)
    do not have four different digits ?

    There are 7000 numbers from 2000 to 8999.

    We'll find the number of numbers that do have four different digits.
    . . The first digit has 7 choices: \{2,3,4,5,6,7,8\}
    . . The other three digits are any permutation of the remaining 9 digits: . P(9,3) \:=\:504
    . . Hence, there are: . 7\cdot504 \:=\:3528 numbers with four different digits.

    Therefore, there are: . 7000 - 3528 \:=\:3472 numbers that do not have four different digits.

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  3. #3
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    For the first one (I hope I don't miss some).

    The number has to be > 5000. for the 4 digit numbers the first digit must be a 5, 6 or 7. We place the 5 in the first spot and arrange 3 of the remaining 4
    digits in P(4,3)=24 ways.

    Do the same for 6 and 7.

    Then for the remaining 5 digit numbers we have 5!=120 ways.

    There are 120+24+24+24=192 numbers.
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  4. #4
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    Thanksssssss!!! I Got It !!
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