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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

Grade:12

2 Answers

Vikas TU
14149 Points
7 years ago
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 .
mycroft holmes
272 Points
7 years ago
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

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