Digits given are 1,2,3,4,5,7,9
Now multiple of 11→ Difference of sum at even & odd place should be divisible by 11.
Let number of the form abcdefg
∴(a+c+e+g)−(b+d+f)=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
(b,d,f)∈(1,2,7)(2,3,5)(1,4,5)
(a,c,e,g)∈(1,4,7,9),(3,4,5,9),(2,3,7,9)
∴ Total number in case-1 =(3!×3)(4!)=432
Case-2
a+c+e+g=10
b+d+f=21
(a,b,e,g)∈1,2,3,4
(b,d,f)&{(5,7,9)}
∴ Total number in case 2=3!×4!=144
∴ Total numbers =144+432=576