The number of 7 -digit numbers which are multiples of 11 and are...

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 $abcdefg$

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

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

$∴$ either $a + c + e + g = 21$ or $10$

$∴ b + d + f = 10$ 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\}$</p><p>\left(a, c, e, g\right) \in \left\{\left(1, 4, 7, 9\right), \left(3, 4, 5, 9\right), \left(2, 3, 7, 9\right)\right\}$

$∴$ 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\}$</p><p>\left(b, d, f\right) \text{ and } \left\{\left(5, 7, 9\right)\right\}$

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

$∴$ Total numbers $= 144 + 432 = 576$