Question

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 _______.

Answer 1

Digits are $1,2,3,4,5,7,9$
Multiple of $11 \rightarrow$ Difference of sum at even $\&$ odd place is divisible by $11$ .
Let number of the form $abcdefg$
$\therefore( a + c + e + g )-( b + d + f )=11 x$
$a+b+c+d+e+f=31$
$\therefore$ either $a + c + e + g =21$ or 10
$\therefore b + d + f =10$ or 21
Case- $1$
$a + c + e + g =21$
$b + d + f =10$
(b, d, f) $\in\{(1,2,7)(2,3,5)(1,4,5)\}$
(a, c, e, g) $\in\{(1,4,7,9),(3,4,5,9),(2,3,7,9)\}$
$\therefore$ Total number in case- $1=(3 ! \times 3)(4 !)=432$
Case- 2
$a + c + e + g =10 $
$b + d + f =21$
$( a , b , e , g ) \in\{1,2,3,4)\}$
$( b , d , f ) \&\{(5,7,9)\} $
$\therefore $ Total number in case $ 2=3 ! \times 4 !=144$
$\therefore $ Total numbers $=144+432=576$