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Math Help - combinatorics

  1. #1
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    combinatorics

    How many strings of five decimal digits
    a. begin with an odd digit?
    my answer)  5.10^3
    b. do not contain the same digit twice?
    my answer) 10.9.8.7.6
    c. have exactly three digits that are 9s?
    my answer)  90

    that's 5.2=10 places for remaining 9 digits
    and 9 choices for each place = 10x9 = 90, but,
    can't it be like this....
    exactly 3 same digits means  9^3.10^2?

    d. contain exactly three distinct digits?
    my answer) 10.9.8.7.10
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  2. #2
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    In a) 5\cdot 10^4

    In b) can the first digit be zero? Such as 06453.
    If yes, then your answer is correct.
    If not then it should 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6
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  3. #3
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    Quote Originally Posted by Plato View Post
    In a) 5\cdot 10^4

    In b) can the first digit be zero? Such as 06453.
    If yes, then your answer is correct.
    If not then it should 9 \cdot 9 \cdot 8 \cdot 7 \cdot 6
    What about c) and d)?
    Can you answer the confusion that I am having with c)?
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  4. #4
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    You did not answer the question about the first digit.
    Can the first digit be zero as in 06236??
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  5. #5
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    Quote Originally Posted by Plato View Post
    You did not answer the question about the first digit.
    Can the first digit be zero as in 06236??
    It hasn't been explicitly mentioned, so I guess 0 counts too.
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  6. #6
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    Assume that the number can have the begin with a zero.

    c. have exactly three digits that are 9s?
    {5 \choose 3} \cdot 9^2.
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  7. #7
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    Quote Originally Posted by Plato View Post
    Assume that the number can have the begin with a zero.

    c. have exactly three digits that are 9s?
    {5 \choose 3} \cdot 9^2.
    What if it cannot begin with a zero? What will it be then?

    Also can you answer the question c) without using "Combination" i.e. by only using "Product Rule"?
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  8. #8
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    {4 \choose 2} \cdot 9^2 + 8 \cdot {4 \choose 3} \cdot 9 if the first is cannot be zero.
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  9. #9
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    Hello, robocop_911!

    Since it says "strings" rather than "five-digit numbers",
    . . I assume they can begin with zero.


    How many strings of five decimal digits

    c. have exactly three digits that are 9's?

    There are {5\choose3} ways to place the 9's.
    The other two places can be filled in 9^2 ways.

    My answer: .  10 \times 81 \:=\:8,100



    d. contain exactly three distinct digits?
    This takes a bit more work . . .

    The string is of the form: . (1)\;aaabc\,\text{ or }\,(2)\;aabbc


    Form (1): . aaabc

    There are {10\choose3} choices for the three digits.
    Then we decide which digit is the "triple": 3 choices.
    And the five digits can be arranged in: . \frac{5!}{3!} ways.

    . . There are: . 120 \times 3 \times 20 \:=\:7,200 strings of the form aaabc.


    Form (2): . aabbc

    There are {10\choose3} choices for the three digits.
    We we decide which two digits are the "pairs": 3 choices.
    And the five digits can be arranged in: . \frac{5!}{<br />
2!2!} ways.

    . . There are: . 120 \times 3 \times 30 \:=\:10,800 strings of the form aabbc.


    Therefore, there are: . 7,200 +10,800\:=\:{\color{blue}18,000} strings with exactly three distinct digits.

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  10. #10
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    Quote Originally Posted by Soroban View Post


    Since it says "strings" rather than "five-digit numbers",
    . . I assume they can begin with zero.



    There are {5\choose3} ways to place the 9's.
    The other two places can be filled in 9^2 ways.

    My answer: .  10 \times 81 \:=\:8,100
    Shouldn't it be P(5,3) since C(5,3) includes 9_1 9_2 9_3
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  11. #11
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    May I point out to you that Soroban's answer is exactly the same that I gave you. Once again, you are showing a real weakness with respect to definitions.
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