The number of 7-digit numbers which are multiples of 11 and are formed using all the digits 1, 2, 3, 4, 5, 7 and 9 is _____________.

Digits are $1,2,3,4,5,7,9$<br/><br/> Multiple of $11 \rightarrow$ Difference of sum at even and odd place is divisible by 11 .<br/><br/> Let number of the form <b>abcdefg</b><br/><br/> $$ \begin{aligned} &\therefore(\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g})-(\mathrm{b}+\mathrm{d}+\mathrm{f})=11 \mathrm{x} \\\\ &\mathrm{a}+\mathrm{b}+\mathrm{c}+\mathrm{d}+\mathrm{e}+\mathrm{f}=31 \\\\ &\therefore \text { either } \mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=21 \text { or } 10 \\\\ &\therefore \mathrm{b}+\mathrm{d}+\mathrm{f}=10 \text { or } 21 \end{aligned} $$<br/><br/> <b>Case-1</b><br/><br/> $$ \begin{aligned} &a+c+e+g=21 \\\\ &b+d+f=10 \\\\ &(b, d, f) \in\{(1,2,7)(2,3,5)(1,4,5)\} \\\\ &(a, c, e, g) \in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\} \end{aligned} $$<br/><br/> <b>Case-2</b><br/><br/> $$ \begin{aligned} &\mathrm{a}+\mathrm{c}+\mathrm{e}+\mathrm{g}=10 \\\\ &\mathrm{~b}+\mathrm{d}+\mathrm{f}=21 \\\\ &(\mathrm{a}, \mathrm{b}, \mathrm{e}, \mathrm{g}) \in\{1,2,3,4)\} \\\\ &(\mathrm{b}, \mathrm{d}, \mathrm{f}) \&\{(5,7,9)\} \\\\ &\therefore \text { Total number in case } 2=3 ! \times 4 !=144 \\\\ &\therefore \text { Total numbers }=144+432=576 \end{aligned} $$