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:

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answer is B.

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Detailed Solution

Concept:

Write the divisibility rule of 11 first .

If the digit is expressed as (

a + c + e) (b + d + f) is 0

or a multiple of 11, then the digit is divisible by 11.
We have options that can satisfy the divisibility of

11 : a, b, c, d, e, and f .

If any value between

7, 5, 0 and a, c, e is possible . B, D, and F

can then choose

9, 2 or 1 .

But a cannot accept 0 because doing so would result in a five-digit number.
has two possibilities, and if one of them is chosen by an or e,
will be given only one choice.
Similar to this, b has three alternatives, d has two options, and f is left with one option.
Total instances for

abcdef

where a, c, and e can take any value between 7, 5, and 0 and b, d, and f can take either

9, 2 or 1 are 2 \times 3 \times 2 \times 2 \times 1 \times 1 = 24

Consider another where b, d, and f can select any value from

7, 5, 0, and a, c, and e

can select either

9, 2 or 1 .

B has three alternatives, d has two, and f is left with one.
Likewise, a has three alternatives, c has two, and e is left with one.
The total number of ways in this example are