Video Solution
https://www.youtube.com/watch?v=X07pj-9KEwM
L I L A C
!1 = 0
!2 = 1
!3 = 2(1+0) = 2
!4 = 3(2+1+ = 9
!5 = 4*(9+2) = 44
As $!n = (n-1)[!(n-1)+ !(n-2)]$
Number of ways we can put each alphabet at different place will be !5 i.e. 44
but we have to consider all the cases below
CASE 1: if first L goes in second L’s shoes that means !4 ways are there as I dont care where the second L is
CASE 2: IF second L goes in First L’s shoes that means !4 ways are there as I dont care where the first L is
CASE 3: if First L and Second L swap there place then we have !3 ways of arranging IAC
All these cases have to be subtracted from !5
And as we have 2 L’s then we did counting twice, so we will divide by 2!
Ans = $\frac{!5-!4-!4-!3}{2!}$
which is $\frac{44-9-9-2}{2}=12$