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mathmath333
 one year ago
How many 6 digit numbers are having 3 odd
and 3 even digits ?
mathmath333
 one year ago
How many 6 digit numbers are having 3 odd and 3 even digits ?

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mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2\(\large \color{black}{\begin{align} & \normalsize \text{ How many 6 digit numbers are having 3 odd}\hspace{.33em}\\~\\ & \normalsize \text{ and 3 even digits ?}\hspace{.33em}\\~\\ & a.)\ 55 \hspace{.33em}\\~\\ & b.)\ (5\cdot 6)^{3}\cdot (4\cdot 6)^{3} \hspace{.33em}\\~\\ & c.)\ 281250 \hspace{.33em}\\~\\ & d.)\ \normalsize \text{none of these} \hspace{.33em}\\~\\ \end{align}}\)

ShadowLegendX
 one year ago
Best ResponseYou've already chosen the best response.0uhhh, (z obviously

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok first you must consider that the first digit can not be zero or it wont be 6 digit number

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0there can be 20 ways odd and even digits can be rearranged

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so from 110 there can 5 odd and 5 even digits are possible

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[5*5*5*5*5*5=15625\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and from 1120 make sure that the first digit cant be zero so

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[4*5*5*5*5*5=12500\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now because we have 5 possible odd and 5 possible even 5+5=10

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Case 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
 one year ago
Best ResponseYou've already chosen the best response.0so 10*15625+10*12500=281250

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so 281250 different numbers with 3 odd and 3 even

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5http://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
 one year ago
Best ResponseYou've already chosen the best response.5Yay, 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
 one year ago
Best ResponseYou've already chosen the best response.5Subtract: http://www.wolframalpha.com/input/?i=%286+choose+3%29*5%5E6++%285+choose+2%29*5%5E5

mathmath333
 one year ago
Best ResponseYou've already chosen the best response.2how this comes \(\dbinom{5}{2}\times5^2 \times 5^3\)

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5If 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
 one year ago
Best ResponseYou've already chosen the best response.5Even if we'd counted the number of ways to choose the odddigit'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
 one year ago
Best ResponseYou've already chosen the best response.2oh i see the irony

ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.5Yeah. 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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