Question

The number of 6 digit numbers that can be formed using the digits 0, 1, 2, 5, 7 and 9 which is divisible by 11 and no digit is repeated is:
\[\begin{align}
  & A.36 \\
 & B.60 \\
 & C.48 \\
 & D.72 \\
\end{align}\]

Answer 1

Hint: In this question, we are given six digits and we have to find the number of ways 6 digit numbers can be formed which are divisible by 11 when no digit is repeated. For this, we will first learn what numbers are divisible by 11 and then use it to find possible cases through which these digits can form a 6 digit number and then calculate total numbers. We will use the divisibility rule of 11 that is, any number whose absolute difference between sum of digits in even position and sum of digits in odd position is 0 or divisible by 11 is itself also divisible by 11.

Complete step by step answer:
Here, we are given six digits as 0, 1, 2, 5, 7, 9. We have to find numbers of six digits using these digits divisibility by 11, when no digit is repeated. As we know, the divisibility rule of 11 states that, any number whose absolute difference between sum of digits in even position and sum of digits in odd position is 0 or divisible by 11 is itself also divisible by 11. So, let us use this rule to find required results.
Sum of given digits is 0+1+2+5+7+9 = 24. As we can see any combination of the sum of three numbers subtracted from the sum of other numbers cannot yield 11 or multiple of 11. So our only option is to get a difference of 0 by considering a pair of sums as 12. If abcdef is our required number then $\left| \left( a+c+e \right)-\left( b+d+f \right) \right|$ should be zero or multiple of 11. Here, it can be zero only, therefore, a+c+e = 12 and b+d+f = 12. Now, let us find ways of taking (a+c+e) and (b+d+f) using the digits 0, 1, 2, 5, 7 and 9.
Case I: Let us suppose a, c, e can take value from 7, 5, 0 then b, d, f will take value from 9, 2 or 1. But ‘a’ cannot take 0 as it will make a 6 digit number. So, 'a' has 2 options, c has 2 options and ‘e’ will have one option left.
Similarly, b can take 3 options, d has 2 options and f has one option. So, total cases becomes $\Rightarrow 2\times 3\times 2\times 2\times 1\times 1=24$.
Case II: Let us suppose a, c, e can take values from 9, 2, 1 then b, d, f will take value from 7, 5, 0. Now, 'a' can have 3 options, c have two options and e has one option. Similarly, b have 3 options, d have two options and f has 1 option.
So, total cases becomes $\Rightarrow 3\times 3\times 2\times 2\times 1\times 1=36$.
Therefore, the total number of 6 digits numbers that can be formed using 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.

So, the correct answer is “Option B”.

Note: Students should know the divisibility rule of 11 for solving this question. Students should take care of all the possibilities before giving a final answer. Students should not forget about not taking zero as the first number as it will result in forming a five digit number.