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

  • 1
    $\begingroup$ Did you mean PENDULUM? PENDULAM is not an English word. Of course, it is a reasonable question for PENDULAM as well which avoids the complication of having two Us. My answer is using PENDULAM $\endgroup$ Commented Feb 20, 2018 at 16:51
  • $\begingroup$ Please read this tutorial on how to typeset mathematics on this site. $\endgroup$ Commented Feb 20, 2018 at 17:18
  • $\begingroup$ I would interpret the phrase "such that vowels are never together" to mean that no 2 vowels are adjacent. $\endgroup$ Commented Feb 20, 2018 at 18:26

1 Answer 1


Your approach 1 deducts only cases where all three vowels are together and does not consider the ordering of the vowels. A correct application of that approach would say there are $8!-6!3!=36000$ ways to order the letters so that all three vowels are not together.

Your approach 2 demands that no pair of vowels be together. It is a correct solution to a different problem.


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