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. 72
B. 60
C. 48
D. 36

Answer 1

Hint: Find the possible values of the numbers that odd places and even places can take. Make all the possible cases. Calculate the total number of ways for each case, then add the results of both the cases to get the required answer.

Complete step-by-step answer:
First of all, write the divisibility rule of 11.
Let the digit is written as $abcdef$, then, the digit is divisible by 11 if and if $\left( {a + c + e} \right) - \left( {b + d + f} \right)$ is 0 or a multiple of 11.
We have possible that $a,b,c,d,e,f$ can take to satisfy the divisibility of 11.
If each of $a,c,e$ can take any value from 7,5,0. Then $b,d,f$ can take 9,2 or 1.
But, $a$ cannot take 0 as it will make a five-digit number then.
$a$ has 2 options, \[c\] has 2 options, if any of them is taken by $a$ and \[e\] will be left with only 1 option.
Similarly, $b$ can take 3 options, $d$ has 2 options and $f$ has one option left.
Total cases are for $abcdef$ when $a,c,e$ can take any value from 7,5,0 and $b,d,f$ can take 9,2 or 1are $2 \times 3 \times 2 \times 2 \times 1 \times 1 = 24$
Similarly take the case when $b,d,f$ can take any value from 7,5,0 and $a,c,e$ can take 9,2 or1.
$b$ can take 3 options, $d$ has 2 options and $f$ has one option left
And similarly $a$ can take 3 options, \[c\] has 2 options and \[e\] has one option left.
Hence, total number of ways for this case is $3 \times 3 \times 2 \times 2 \times 1 \times 1 = 36$
Therefore, 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 sum of 24 and 36, which is 60.
Hence, option 2 is correct.

Note: Students must know the divisibility rule of 11. The divisibility rule of 11 states that the difference of sum of odd places and sum of even places must be a multiple of 11 or should be 0.