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```
number of ways we can arrange 1,2,3,4,...8 such that the product of any five consecutive position is divisible by 5
number of ways we can arrange 1,2,3,4,...8 such that the product of any five consecutive position is divisible by 5

```
4 years ago

Vikas TU
14149 Points
```							Hint: First find out the possible ways of product of five numbers havinng last digit 0 or 5 .u will have to check this first.then find the consecutiveness order .
```
4 years ago
mycroft holmes
272 Points
```							If 5 is placed in the 4th or 5th place, then the product of any 5 consecutively placed numbers will be divisible by 5. So, for 5, we have two choices of placement, and corresponding to each, the other digits can be permuted in 7! ways. Hence, total number of ways = 2 X 7! = 10080
```
4 years ago
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