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

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

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

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