Question

(c) In how many way we can list the digits{1,1,2,2,3,4,5}so that notwo identical digit are in consecutive positions?

Answer 1

660 ways

Explanation:

we have two numbers in consecutive positions in this question

(1,1) and (2,2)

numbers of ways that (1,1) are in consecutive positions = 6!/2! = 360

number of ways that (2,2) are in consecutive positions = 6!/2! = 360

the permutation of (11),(22),3,4,5 = 5!

ps: I counted the pairs as one each.

5! = 120

to get total number of permutations

7!/2!2!

= 5040/4

= 1260

the number of ways that 2 identical digits are not consecutively positioned = 1260-360-360+120

= 660 ways