A counting problem

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?

My solution:

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

Could someone please check if my solution is correct? Thanks.

Quote:

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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

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That is correct.