Find the number of permutations of the letters of the word ‘PENDULAM', such that vowels are never together.
(Approach 1)I know one solution is to find the permutation that vowels are together (ie; 6!) and subtract it from total number of permutations for the word (ie; 8!).
Approach 2: I came across another interesting approach like shown below :
Consonants in the word = $5$ ie;{P,N,D,L,M}
Ways of arranging $5$ letters = $5!$
Consider a case like PNDLM and we insert positions that satisfy the condition of vowels not being together like _P_N_D_L_M_.
So the vowels can be in any of the $6$ dashed positions (which is a equivalent to choosing $3$ out of $6$ positions ie; $6C3$) and vowels for chosen $3$ positions can be arranged among themselves in $3!$ ways. So the number of permutations should be : $6C3 \cdot 3! \cdot 5! = 14400$ (using counting principle)
whereas solution by Approach 1 would be $= 8! - 6! = 39600$.
Can someone point out where Approach 2 went wrong. I would like to use Approach 2 so that I can solve cases like
permutation of $2$ vowels not being together