Help plz with counting
Find the number of ways of arranging the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 in a row so that the product of any two adjacent numbers is even or a mutliple of 3
There are: 10 P 2==90 2-digit permutations
(1, 5) ,(1, 7) ,(5, 1) ,(5, 7) ,(7, 1) ,(7, 5) ,>>Total = 6
The above 6 permutations of {1, 5, 7}, in groups of 2, when multiplied together, are neither divisible by 2 or by 3.
As a result, when they are included with remaining 8 digits in a 10-digit permutation, give a total of: 6 x 8==48 "bad" permutations, which must be exlcluded from the total of 10! permutations.
And we can do that by using proportionality: That is: 48 / 90 x 10! ==1,935,360 "bad" permutations. And:
10! - 1,935,360 ==1,693,440 - "good" permutations possible.
