Question

The number of words formed from letters of the word ‘EAMCET’ so that no two vowels come together.

Answer 1

As we know that in this question it is very difficult or impossible to count the words by ourselves. The method for solving these kinds of questions is by permutations and combinations.

Given: The word ‘EAMCET’ is being given. It has 3 consonants and 3 vowels, but, since ‘E’ is repeated. Therefore, the number of vowels will be 2. No two vowels should come together to find out this question.

Complete step-by-step answer:
Let us arrange 3 consonants first. This can be done in \[^3{P_3}\] ways i.e. 3 ways. mark these all consonants by C as shown in the following figure:

Now no two vowels are together, if these are arranged only at the cross marked placed (‘X’). These are 4 cross marked places and the 2 vowels (Totally 3 but since E is repeating, therefore 2 vowels) which can be arranged at these places in \[^4{P_3}\] ways. Hence, by multiplication principle of counting, the numbers of ways of arranging the letters of the word ‘EAMCET’ so that no two vowels are together will be

\[
   = \left| \!{\underline {\,
  3 \,}} \right. { \times ^4}{P_2} \\
   = (3 \times 2 \times 1) \times \dfrac{{4 \times 3 \times 2 \times 1}}{{2 \times 1}} \\
   = 6 \times 12 \\
   = 72 \\
\]

Hence, the number of words formed by the letter of the word ‘EAMCET’
With no vowels comes together = 72 words.

Note: In these types of questions we must have studied permutations and combinations, which are used in these types of little bit typical questions. In these questions \[\left| \!{\underline {\,
  x \,}} \right. \] is factorial of x while \[^x{P_y}\] is equal to factorial of x divided by factorial of (x-y). Try to visualise the problem with the help of small diagrams.