Find number of different permutations of all the letters of the word "PERMUTATION" such that any two consecutive letters are neither both vowels nor both identical.

The vowels in this word are A,E,I,O,U and identical letters are T,T

I tried in two ways:

Method $1.$ Number of arrangements of the letters P,R,M,N,T,T such that two T's do not occur together is given by

$$\frac{6!}{2!}-5!=2 \times 5!$$ Now in each of these arrangement there are 7 gaps whch can be occupied by $5$ vowels for which the number of ways is $$\binom{7}{5} \times 5!=21 \times 5!$$ Hence total number of ways is

$$ 42 \times 5! \times 5!$$

Method $2.$ All possible permutations of PRMNTT is $$\frac{6!}{2!}$$ and there are $7$ gaps and no of ways which can be occupied by $5$ vowels is $$\binom{7}{5} \times 5!$$ so total number of ways is $$\frac{6!}{2!} \times \binom{7}{5} \times 5!=63 \times 5! \times 5! \tag{A}$$.

But among these, number of words in which two T's occur together(assuming TT as one letter) is given by

$$\binom{6}{5} \times 5! \times 5!=6 \times 5! \times 5! \tag{B}$$

Hence to total number of ways is $(A)-(B)$ which comes out as $$57 \times 5! \times 5!$$

can some one help me why i am getting different answer,in which method i went wrong?

  • $\begingroup$ Assuming $TT$ is one letter say $X$, number of ways $P,R,M,N,X$ can be arranged is $5!$ and there are $6$ gaps to be occupied by $5$ vowels which can be done in $\binom{6}{5} \times 5!$. multiplying both we get formula $(B)$ $\endgroup$ – Ekaveera Kumar Sharma Dec 16 '15 at 11:06
  • 1
    $\begingroup$ The problem in your set-up is that the placement of the $T$'s can cause gaps of size $1$ to appear and these gaps can be filled by a vowel without violating the rules. $\endgroup$ – Michael Burr Dec 16 '15 at 11:13

The error is in method 1. Indeed first you counted the number of permutations of the consonants excluding the case when TT appears and then you insert the vowels. You forgot that a vowel can be inserted between TT, creating a valid permutation.

Method 2 seems correct and you can notice that you got a greater number precisely because you counted the cases you missed in method 1.

  • $\begingroup$ Thanks it was pretty clear now $\endgroup$ – Ekaveera Kumar Sharma Dec 16 '15 at 12:03

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