Question

How many eight digit no can be formed using $1,1,2,2,3,3,4,5$ if no $2$ consecutive digits are identical.

Answer 1

Odd digits cannot occupy odd places,so odd digits should occupy even places i,e 4 possiblities and odd numbers are three.
1) So selecting three positions from 4 positions can be done in 4C3 ways that is 4 ways.
2) Filling the three positions with three numbers can be done is 3! ways but 1,1 are identical so total number of ways is 3!/2! i,e 3
Now filling the remaining 5 positions can be done in 5! ways but again 2,2,2 are identical and 4,4 are identical so number of ways is 5!/(3!*2!)
i,e in 10 ways so the total number of ways is
4*3*10=120.