+188 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.)
+6251 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.
+6251 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.
