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mathmath333

  • one year ago

How many 6 digit numbers are having 3 odd and 3 even digits ?

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  1. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align} & \normalsize \text{ How many 6 digit numbers are having 3 odd}\hspace{.33em}\\~\\ & \normalsize \text{ and 3 even digits ?}\hspace{.33em}\\~\\ & a.)\ 55 \hspace{.33em}\\~\\ & b.)\ (5\cdot 6)^{3}\cdot (4\cdot 6)^{3} \hspace{.33em}\\~\\ & c.)\ 281250 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

  2. ShadowLegendX
    • one year ago
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    uhhh, (z obviously

  3. anonymous
    • one year ago
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    ok first you must consider that the first digit can not be zero or it wont be 6 digit number

  4. anonymous
    • one year ago
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    there can be 20 ways odd and even digits can be rearranged

  5. anonymous
    • one year ago
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    so from 1-10 there can 5 odd and 5 even digits are possible

  6. anonymous
    • one year ago
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    \[5*5*5*5*5*5=15625\]

  7. anonymous
    • one year ago
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    and from 11-20 make sure that the first digit cant be zero so

  8. anonymous
    • one year ago
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    \[4*5*5*5*5*5=12500\]

  9. anonymous
    • one year ago
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    now because we have 5 possible odd and 5 possible even 5+5=10

  10. ParthKohli
    • one year ago
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    Case 1: Numbers that do not contain zero. \(\binom{6}{3}\) ways to choose the place for even numbers. \(4\) choices for each blank for even and \(5\) choices for each blank for odd. \(\binom{6}{3}\times4^3 \times 5^3\) such numbers in this case. Case 2: Numbers that contain zero. 2.1: Containing one zero. \(\binom{5}{1}\) ways to choose zero's place. \(\binom{5}{2}\) ways for the other two even numbers. Total \(\binom{5}{1}\times \binom{5}{2}\times 4^2\times 5^3 \) 2.2: Containing two zeroes. \(\binom{5}{2}\) ways to choose zero's place. \(\binom{4}{1}\) ways to choose the last even number's place. Total \(\binom{5}{2} \times \binom{4}{1} \times 4 \times 5^3 \) 2.3: Containing three zeroes. \(\binom{5}{3}\) ways to choose zero's place. Total \(\binom{5}{3}\times 5^3\)

  11. anonymous
    • one year ago
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    so 10*15625+10*12500=281250

  12. anonymous
    • one year ago
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    so 281250 different numbers with 3 odd and 3 even

  13. ParthKohli
    • one year ago
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    Yay, I like that approach too. All numbers: \(\binom{6}{3}\times 5^3 \times 5^3\) Numbers starting with zero: \(\binom{5}{2}\times5^2 \times 5^3 \)

  14. ParthKohli
    • one year ago
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    Subtract: http://www.wolframalpha.com/input/?i=%286+choose+3%29*5%5E6+-+%285+choose+2%29*5%5E5

  15. mathmath333
    • one year ago
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    how this comes \(\dbinom{5}{2}\times5^2 \times 5^3\)

  16. ParthKohli
    • one year ago
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    If a number starts with zero... 0 _ _ _ _ _ Now there are two even digits and three odd digits left. There are \(\binom{5}{2}\) ways to choose the even digit's place as there are five blanks left and two even digits left. Now there are \(5\) even digits (0, 2, 4, 6, 8) and \(5\) odd digits (1, 3, 5, 7, 9) so I just multiplied that.

  17. ParthKohli
    • one year ago
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    Even if we'd counted the number of ways to choose the odd-digit's place, i.e., \(\binom{5}{3}\), nothing would have changed because \(\binom{5}{3} = \binom{5}{2}\) and that is the power of combinatorics. :)

  18. mathmath333
    • one year ago
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    oh i see the irony

  19. ParthKohli
    • one year ago
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    Yeah. The second approach is better in an exam situation, but the first approach (building up from zero) is one that you can always use even if nothing strikes you.

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