Question

If as many more words as possible be formed out of the letters of the word “DOGMATIC” then find the number of words in which the relative order of vowels and consonants remain unchanged.

Answer 1

In order to find solution to this problem, we will have to first find out number of vowels and number of consonants from our word “DOGMATIC” and then find the number of words so that the relative order of vowels and consonants remain unchanged with the help of permutation and combination concepts and the product rule.

Complete step by step solution:
Consider the word given “DOGMATIC”.
Now, we have to find the total number of vowels and consonants in our word “DOGMATIC”.
Therefore, the total number of vowels = O, A, I $=3$
And, the total number of consonants = D, G, M, T, C $=5$
Therefore, there are $3$ vowels and $5$ consonants in the word DOGMATIC.
Now as we have $3$ vowels which we can arrange as $3!$ ways and $5$ consonants which we can arrange as $5!$ ways. Also, we have to keep relative order of vowels and consonants unchanged.
So, the total number of words that can be formed so that relative position of vowels and consonants remain unchanged can be:
So by using product rule of permutation and combination, we get:
$=3!\times 5!$
On simplifying, we get:
$=6\times 120$
On simplifying, we get:
$=720$
Now, since the question says, “as many more words as possible”. $720$ is the count including DOGMATIC, therefore new words formed will be:
$= 720-1$
That is, we get:
$= 719$
Therefore, total words possible will be $719$.

Note: Whenever we get this type of question the basic way of solving is explained but we know directly that if we have n objects and we have to fill ‘n’ spaces then the total number of ways will be $n!$ and here product rule is used so we should have knowledge about this rule.
The Product Rule is: If there are n(A) ways to do A and n(B) ways to do B, then the number of ways to do A and B is $n\left( A \right)\times n\left( B \right)$.