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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?

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

This can be done in $^6C_2$ ways.
This can be done in $2^4$ ways
$\therefore$ The required number of 6 digit numbers= $2^4.^6C_2$