Question

How many permutations of 11223344 have no equal numbers next to each other?

Answer 1

Let Si be the set of all the permutations which contain an adjacent pair ii and let S be the set of all possible permutations on 1,1,2,2,3,3,4,4

Hence the required permutations = |S|−|S1∪S1∪S3∪S4|

|S1∪⋯∪S4|=∑N(i)−∑N(i,j)+∑N(i,j,k)−∑N(i,j,k,l)

N(i)=7!/2³

Where N(1) is number of permutations of the 8-element set (11),2,2,3,3,4,4 where we treat the adjacent 1s as a single element.

Similarly,

N(i,j)=6!/2²

N(i,j,k)=5!/2

N(i,j,k,l)=4!

Hence, Required Permutation = 8!/2⁴ -⁴C₁*7!/2³ + ⁴C₂* 6!/2² - ⁴C₃* 5!/2 + ⁴C₄* 4! = 864