answer is right there in the strings: S = PPXXPXPXPPX...
Possible permutations = No. of ways to generate strings of P/X's such that
in each sub-string S[0..i] no. of P's is always greater than or equal to
no. of X's.
You can also view this as - all possible strings of n balanced parentheses.
This
we can generate permutations using stack. For example, if n=1,2,3 and
Push() is denoted by "P", and Pop() by "X",
then we can generate permutations as :
PPPXXX = 321
PPXXPX = 213
PXPXPX = 123
PXPPXX = 132
PPXPXX = 231
(note : at any point, #P's>=#X's and finally, #P's=#X's)
ques :- Given n eleme
