# A counting problem

• Dec 19th 2011, 01:27 AM
alexmahone
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.
• Dec 19th 2011, 05:19 AM
Plato
Re: A counting problem
Quote:

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

That is correct.