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How many 6 digit numbers can be formed using the digits $1,2,3$ so that the digit 1 occur 2 times in each number?

$\begin{array}{1 1} 2^4.^4C_2 \\ ^6C_2 \\ 4^2.^6C_2 \\ 2^4.^6C_2 \end{array}$

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There are 6 places in 6 digit numbers.
Out of these 7 places two places are to be fixed for number 1.
This can be done in $^6C_2$ ways.
The remaining 4 places can be fixed by any of the two digits 2,3.
This can be done in $2^4$ ways
$\therefore$ The required number of 6 digit numbers= $2^4.^6C_2$
answered Aug 15, 2013 by rvidyagovindarajan_1

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