Originally Posted by

**alexmahone** In how many ways can we list the digits {1,1,2,2,3,4,5} so that two identical digits are not in consecutive positions?

No. of ways in which the 1's are in consecutive positions = 6!/2! = 360

No. of ways in which the 2's are in consecutive positions = 6!/2! = 360

No. of ways in which both 1's and 2's are in consecutive positions = 5! = 120

No. of ways in which either the 1's or 2's are in consecutive positions = 360+360-120=600

Total number of permutations = 7!/(2!2!) = 1260

No. of ways in which two identical digits are not in consecutive positions = 1260-600=660