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Q.
Number of 6 digit numbers divisible by 11 made by using digits 0, 1, 2, 5, 7, 9 without repetition is equal to
Solution:
Find the total number of ways
Given digits are $0,1,2,5,7$ and 9
The number is said to be divisible 11 if the difference between the sum of digits at the odd and even places equ 0 or divisible by 11
Let the six digit number be = xyzuvw
The difference of $(y+u+w)-(x+z+v)=0$ or divisible by 11
Using the above digits
Case - I $\{ x , z , v \}=\{9,2,1\},\{ b , d , f \}=\{7,5,0\}$
Numbers can be formed $=3 ! \times 3 !=36$
Case - II $\{ a , c , e \}=\{7,5,0\},\{ b , d , f \}=\{9,2,1\}$
Numbers can be formed $=2 \times 2 ! \times 3 !=24$
$\therefore$ The total numbers can be formed $=24+36=60$