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How many sequences of 6 digits x1, x2, ... , x6 can we form, given the condition that no two adjacent xi have the same parity? Leading zeroes are allowed. (Parity means 'odd' or 'even'; so, for example, x2 and x3 cannot both be odd or both be even.)

 Dec 8, 2018

Best Answer 

 #1
avatar+5222 
+1

well it's clear that among your x's 3 must be even and 3 must be odd.

 

The evens can occupy either slots 1,3,5 or slots 2,4,6

 

There are 3! = 6 ways you can arrange them in those slots.

 

The odds will occupy the remaining slots and again there are 6 ways to arrange them.

 

So in total there are 6*6=36 ways to arrange your x's as described.

 Dec 8, 2018
 #1
avatar+5222 
+1
Best Answer

well it's clear that among your x's 3 must be even and 3 must be odd.

 

The evens can occupy either slots 1,3,5 or slots 2,4,6

 

There are 3! = 6 ways you can arrange them in those slots.

 

The odds will occupy the remaining slots and again there are 6 ways to arrange them.

 

So in total there are 6*6=36 ways to arrange your x's as described.

Rom Dec 8, 2018

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