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The number of permutations 1,2,3..8 taken all at a time so that the product of two consecutive digits is even must be-

The number of permutations 1,2,3..8 taken all at a time so that the product of two consecutive digits is even must be-

Grade:12

1 Answers

Riddhish Bhalodia
askIITians Faculty 434 Points
8 years ago
For product of consecutive digits to be even one of thenumbers every alternate place should be even. So the cases wher the product of consecutive digits is not even when there exist two odd numbers together. So we count number of ways where atleast two odd digits are next to one another, and subtract it from total permutations.
Total permutations = 8!
Way to select 2 odd number = ^4C_2
permutation with selected numbers together = 7!.2!
Hence
number of ways where atleast two odd digits are next to one another = ^4C_2.7!.2! – ways both odd numbers are consecutive(to prevent double counting)
= ^4C_2.7!.2! - ^4C_2.2!2!6!
Thus answer = 8! - ^4C_2.7!.2! + ^4C_2.2!2!6!

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