Mathematics - Permutation & Combination Question with Solution | TestHub

Digits given are $1,2,3,4,5,7,9$

Now multiple of $11\to$ Difference of sum at even & odd place should be divisible by $11$.

Let number of the form $abcdefg$

$\therefore \left(a+c+e+g\right)-\left(b+d+f\right)=11x$

$a+b+c+d+e+f=31$

$\therefore$ either $a+c+e+g=21$ or $10$

$\therefore b+d+f=10$ or $21$

Case-1

$a+c+e+g=21$

$b+d+f=10$

$\left(b,d,f\right)\in \left\{\left(1,2,7\right)\left(2,3,5\right)\left(1,4,5\right)\right\}$

$\left(\mathrm{a},\mathrm{c},\mathrm{e},\mathrm{g}\right)\in \left\{\left(1,4,7,9\right),\left(3,4,5,9\right),\left(2,3,7,9\right)\right\}$

$\therefore$ Total number in case-1 $=\left(3!\times 3\right)\left(4!\right)=432$

Case-2

$a+c+e+g=10$

$b+d+f=21$

$\left(a,b,e,g\right)\in \left\{1,2,3,4\right\}$

$\left(b,d,f\right) \& \left\{\left(5,7,9\right)\right\}$

$\therefore$ Total number in case $2=3!\times 4!=144$

$\therefore$  Total numbers $=144+432=576$