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Math Help - [SOLVED] permutations 2

  1. #1
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    [SOLVED] permutations 2

    Find the number of 6 digit numbers that can be formed using the digits 0 to 9 if:

    a) each digit number be different
    b) digits may be repeated
    c) at least one 5 is included
    d) the number is even,greater than or equal to 600,000, with repetition of digits allowed
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  2. #2
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    Hello, william!

    I will assume that the number may not begin with 0 (zero).



    Find the number of 6-digit numbers that can be formed using the digits 0 to 9 if:

    a) each digit must be different
    The first digit cannot be zero: 9 choices.

    The second digit can be any of the other 9 other digits.
    The third digit can be any of the other 8 digits.
    The fourth digit can be any of the other 7 digits.
    The fifth digit can be any of the other 6 digits.
    The sixth digit can be any of the other 5 digits.

    Answer: . 9 \times 9 \times 8 \times  7 \times 6 \times 5 \:=\:130,\!080



    b) digits may be repeated
    The first digit has 9 choices.
    Each of the other five digits has 10 choices.

    Answer: . 9 \times 10^5 \:=\:900,\!000



    c) at least one 5 is included
    In part (b), we saw that there are 900,\!000 possible numbers.


    Consider the numbers that have no 5's.

    The first digit has 8 choices (not 0 and not 5).
    Each of the other five digits has 9 choices.
    Hence, there are: . 8 \times 9^5 \:=\:472,392 numbers with no 5's.

    Answer: . 900,\!000 - 472,\!392 \:=\:427,608




    d) the number is even, greater than or equal to 600,000, with repetition allowed
    The number is \geq 600,\!000
    The first digit must be: 6, 7, 8, or 9 . . . 4 choices.

    The number is even.
    The last digit must be: 0, 2, 4, 6, 8 . . . 5 choices.

    Each of the other four digits has 10 choices.

    Answer: . 4 \times 5 \times 10^4 \:=\:200,\!000

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  3. #3
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    Quote Originally Posted by william View Post
    Find the number of 6 digit numbers that can be formed using the digits 0 to 9 if:

    a) each digit number be different
    You can choose any of the 10 digits to be first, any of the remaining 9 digits to be next, any of the remaining 8 digits to be next, ..., down to any of the remaining 5 remaining digits to be the last, 6th digits. There are 10*9*8*7*6*5= 10!/(10-4)!= 9!/4! such digits.

    b) digits may be repeated
    Now you can choose any of 10 digits to be first, any of those same 10 digits to be next, etc. There are 10^6 such digits.

    c) at least one 5 is included
    If digits can also be repeated, which I assume since you say "at least one 5 is included", then you can do this by selecting a 5 first, then any of 10 digits for the remaining 5 digits: that is 10^5. Since that single 5 can be in any of the 6 places, the total number of such numbers is 6(10^5).

    d) the number is even,greater than or equal to 600,000, with repetition of digits allowed
    The first digit must be 6, 7, 8, 9, four choices and the last digit must be one of 0, 2, 4, 6, 8, five choices. The other 4 digits can be any of the 10 digits. There are 4(5)(10^5)= 2000000 such numbers.

    As usual Soroban got in ahead of me. Blast him!
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, william!

    I will assume that the number may not begin with 0 (zero).



    The first digit cannot be zero: 9 choices.

    The second digit can be any of the other 9 other digits.
    The third digit can be any of the other 8 digits.
    The fourth digit can be any of the other 7 digits.
    The fifth digit can be any of the other 6 digits.
    The sixth digit can be any of the other 5 digits.

    Answer: . 9 \times 9 \times 8 \times  7 \times 6 \times 5 \:=\:130,\!080



    The first digit has 9 choices.
    Each of the other five digits has 10 choices.

    Answer: . 9 \times 10^5 \:=\:900,\!000



    In part (b), we saw that there are 900,\!000 possible numbers.


    Consider the numbers that have no 5's.

    The first digit has 8 choices (not 0 and not 5).
    Each of the other five digits has 9 choices.
    Hence, there are: . 8 \times 9^5 \:=\:472,392 numbers with no 5's.

    Answer: . 900,\!000 - 472,\!392 \:=\:427,608




    The number is \geq 600,\!000
    The first digit must be: 6, 7, 8, or 9 . . . 4 choices.

    The number is even.
    The last digit must be: 0, 2, 4, 6, 8 . . . 5 choices.

    Each of the other four digits has 10 choices.

    Answer: . 4 \times 5 \times 10^4 \:=\:200,\!000

    thank you! for a) you mean 136,080, right?
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