Question

5. ABC-Free Permutations

Consider permutations of the 26-character lowercase alphabet

Σ={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}.

1. In how many of these permutations do a,b,c occur consecutively and in that order?
2. In how many of these permutations does a appear before b and b appear before c?

Answer 1

Σ={a,b,c,d,e,f,g,h,i,j,k,l,m,n,o,p,q,r,s,t,u,v,w,x,y,z}
(a) a,b,c occur consecutively and in that order
consider abc as one character therefore we have total character = 26-3+1 = 24
Permutation = 24!

(b) a appear before b and b appear before c
Select any three places for a, b and c. They need no arrangement amongst themselves as a would always come before b and b would always come before c.
The remaining (26-3) = 23 character have to be arranged in 23 places.
Thus, ²⁶C₃ x 23! = 26! / 3!