mathmath333
  • mathmath333
How many 6 digit numbers are having 3 odd and 3 even digits ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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mathmath333
  • mathmath333
\(\large \color{black}{\begin{align} & \normalsize \text{ How many 6 digit numbers are having 3 odd}\hspace{.33em}\\~\\ & \normalsize \text{ and 3 even digits ?}\hspace{.33em}\\~\\ & a.)\ 55 \hspace{.33em}\\~\\ & b.)\ (5\cdot 6)^{3}\cdot (4\cdot 6)^{3} \hspace{.33em}\\~\\ & c.)\ 281250 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)
ShadowLegendX
  • ShadowLegendX
uhhh, (z obviously
anonymous
  • anonymous
ok first you must consider that the first digit can not be zero or it wont be 6 digit number

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anonymous
  • anonymous
there can be 20 ways odd and even digits can be rearranged
anonymous
  • anonymous
so from 1-10 there can 5 odd and 5 even digits are possible
anonymous
  • anonymous
\[5*5*5*5*5*5=15625\]
anonymous
  • anonymous
and from 11-20 make sure that the first digit cant be zero so
anonymous
  • anonymous
\[4*5*5*5*5*5=12500\]
anonymous
  • anonymous
now because we have 5 possible odd and 5 possible even 5+5=10
ParthKohli
  • ParthKohli
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 \(\binom{5}{3}\times 5^3\)
anonymous
  • anonymous
so 10*15625+10*12500=281250
anonymous
  • anonymous
so 281250 different numbers with 3 odd and 3 even
ParthKohli
  • ParthKohli
http://www.wolframalpha.com/input/?i=20*4%5E3*5%5E3+%2B+5*%285+choose+2%29*4%5E2*5%5E3+%2B+%285+choose+2%29*%284+choose+1%29*4*5%5E3+%2B+%285+choose+3%29*5%5E3 Indeed - 281250 is what I get too.
ParthKohli
  • ParthKohli
Yay, I like that approach too. All numbers: \(\binom{6}{3}\times 5^3 \times 5^3\) Numbers starting with zero: \(\binom{5}{2}\times5^2 \times 5^3 \)
ParthKohli
  • ParthKohli
Subtract: http://www.wolframalpha.com/input/?i=%286+choose+3%29*5%5E6+-+%285+choose+2%29*5%5E5
mathmath333
  • mathmath333
how this comes \(\dbinom{5}{2}\times5^2 \times 5^3\)
ParthKohli
  • ParthKohli
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.
ParthKohli
  • ParthKohli
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. :)
mathmath333
  • mathmath333
oh i see the irony
ParthKohli
  • ParthKohli
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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