Question

The number of $6$ digit numbers that can be formed using the digits $0, 1, 2, 5, 7$ and $9$ which are divisible by $11$ and no digit is repeated, is :

• A. 36
• B. 60
• C. 48
• D. 72

Answer 1

Answer: (B)

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 o. 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 $=24+36=60$