Let S be the set of all those 2007-digit integer of the form $2 a_{1} a_{2} a_{3} ...... a_{2006}$ which contain digit 9 odd number of times in each sequence $a_{1}, a_{2}, a_{3}, ......, a_{2006}$ . The cardinal number of S is

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$\frac{1}{2} (10^{2006} + 8^{2006})$

$10^{2006} + 8^{2006}$

$10^{2006} - 8^{2006}$

$\frac{1}{2} (10^{2006} - 8^{2006})$

answer is B.

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Detailed Solution

Define A as the number of the elements in S, we have
$A = (\begin{matrix} 2006 \\ 1 \end{matrix}) 9^{2005} + (\begin{matrix} 2006 \\ 3 \end{matrix}) 9^{2003} + ..... + (\begin{matrix} 2006 \\ 2005 \end{matrix}) 9$
On the other hand, ${(9 + 1)}^{2006} = \sum_{k = 0}^{2006} (\begin{matrix} 2006 \\ k \end{matrix}) 9^{2006 - k} a n d {(9 - 1)}^{2006} = \sum_{k = 0}^{2006} (\begin{matrix} 2006 \\ k \end{matrix}) {(- 1)}^{k} 9^{2006 - k}$
So, $A = (\begin{matrix} 2006 \\ 1 \end{matrix}) 9^{2005} + (\begin{matrix} 2006 \\ 3 \end{matrix}) 9^{2003} + ..... + (\begin{matrix} 2006 \\ 2005 \end{matrix}) 9 = \frac{1}{2} 10^{2006} - 8^{2006}$