Permutations and Combinations Part 1

Digits are 1,2,3,4,5,7,9
Multiple of 11→ Difference of sum at even backslash& odd place is 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)}
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