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Manan Solanki
:
Find the number of numbers that can be 1,2,3,4,3,2,1 by placing the odd digit at odd places
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December 8, 2016 at 6:51pm
Little Sky
:
This question is asking the total number of combination for which odd digits only occupy odd places. The slots are as shown below in order:[deqn]\underbrace{p1}_{odd}~\underbrace{p2}_{even}~\underbrace{p3}_{odd}~\underbrace{p4}_{even}~\underbrace{p5}_{odd}~\underbrace{p6}_{even}~\underbrace{p7}_{odd}[/deqn]Now just for the odd places:[deqn]\underbrace{p1}_{odd}~\underbrace{p3}_{odd}~\underbrace{p5}_{odd}~\underbrace{p7}_{odd}[/deqn]We have 4 places for 2 of the '1's and 2 of the '3's. By choosing 2 places out of 4 places for the '1's, we have:[deqn]\left( \begin{array}{c} 4 \\ 2 \end{array} \right)[/deqn]We have 2 places left for the '3's. By choosing 2 places out of last 2 places for the '3's, we have:[deqn]\left( \begin{array}{c} 2 \\ 2 \end{array} \right)[/deqn]Just for the odd places, we have:[deqn]\left( \begin{array}{c} 4 \\ 2 \end{array} \right)\left( \begin{array}{c} 2 \\ 2 \end{array} \right)=6\times1=6[/deqn]Now for even places [deqn]\underbrace{p2}_{even}~\underbrace{p4}_{even}~\underbrace{p6}_{even}[/deqn]We have 3 places for 2 of the '2's and 1 of the '1'. By choosing 2 places out of 3 places for the '2's, we have:[deqn]\left( \begin{array}{c} 3 \\ 2 \end{array} \right)[/deqn]We only have 1 place left for the '1'. So, obviously,[deqn]\left( \begin{array}{c} 1 \\ 1 \end{array} \right)=1[/deqn]Just for the even places, we have:[deqn]\left( \begin{array}{c} 3 \\ 2 \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \end{array} \right)=3\times1=3[/deqn]Putting everything together, we have:[deqn]\left( \begin{array}{c} 4 \\ 2 \end{array} \right)\left( \begin{array}{c} 2 \\ 2 \end{array} \right)\left( \begin{array}{c} 3 \\ 2 \end{array} \right)\left( \begin{array}{c} 1 \\ 1 \end{array} \right)=6\times1\times3\times1=18[/deqn]
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December 9, 2016 at 1:03am
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