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 given are $1,2,3,4,5,7,9$1,2,3,4,5,7,9

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

Let number of the form $abcdefg$abcdefg

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

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

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

$\therefore b+d+f=10$∴b+d+f=10 or $21$21

Case-1

$$a+c+e+g=21$$a+c+e+g=21

$$b+d+f=10$$b+d+f=10

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

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

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

Case-2

$$a+c+e+g=10$$a+c+e+g=10

$$b+d+f=21$$b+d+f=21

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

(b,d,f)&{(5,7,9)}

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

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