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# How many 10 digit numbers are there using the digits 1 and 2 ?

$\begin{array}{1 1} (A)\;^{10}C_1+^9C_1 \\ (B) \;2^{10} \\ (C) ^{10}C_2 \\ (D) 10! \end{array}$

Each place can be fixed by any of the two digits.
$\therefore$The required 10 digit numbers = $2^{10}$