Question

How many 8 digit no.S can be formed using 1,1,2,2,3,3,4,5.So that no two same no. Are consecutive?

Answer 1

One way is to use Inclusion/Exclusion. You attempted a form of it, but something more elaborate is needed.

You know that there are 8!2!2!2! possible sequences if we have no restriction.

Now let us count the number that have the two 1's together. Tie them together, and think of them as a superletter. Now we have 7 "letters." These can be arranged in 7!2!2! ways. We get the same count for the 2's together, and for the 3's together. So our first estimate for the required number is 8!2!2!2!−3⋅7!2!2!.

But we have subtracted too much, for we have subtracted one too many times the patterns that have two 1's and two 2's together, as well as the ones in which we have two 2's and two 3's together, and so on. There are 6!2! of each of these, so our next estimate is8!2!2!2!−3⋅7!2!2!+3⋅6!2!.

But we have added back too much, for we have added back one too many times the 5! sequences in which the 1's are together, and the 2's, and the 3's. Our final count is 8!2!2!2!−3⋅7!2!2!+3⋅6!2!−5!.