Question

The number of words which can be made from letters of the word INTERMEDIATE is
A.907200 if the word start with I and end with E
B.21600 if vowels and constraints occupy their originally places
C.43200 if vowels and consonants occur alternatively
D.332400 if all the vowels occur together.

Answer 1

Hint: All the conditions listed in the option are different. Therefore, we will find the number of words formed using each condition. We will use permutations to compute the number of words formed after rearranging the letters of INTERMEDIATE. If letters are repeated, we use the formula, $\dfrac{{n!}}{{{P_1}!{P_2}!.....}}$ , $n$ is the number of letters formed, ${P_1}$ the number of times a letter is repeated, ${P_2}$ the number of times second letter is repeated and so on.

Complete step-by-step answer:
Since, the letters are rearranged to form different words, we will use permutation.
First, let us find the total number of words formed using the word INTERMEDIATE if the words start with I and ends with E.
Fix I and E at first and last positions respectively. We will apply permutations to the rest of the numbers.
Now, there are 10 letters in which T is repeated 2 times, E is repeated 2 times.
Hence, the number of words formed can be calculated using the formula, $\dfrac{{n!}}{{{P_1}!{P_2}!.....}}$.
On substituting the values in the formula, we get, $\dfrac{{10!}}{{2!2!}}$
Solve the expression using the fact that \[n! = n.\left( {n - 1} \right).\left( {n - 2} \right)........3.2.1\]
$\dfrac{{10!}}{{2!2!}} = \dfrac{{10.9.8.7.6.5.4.3.2.1}}{{\left( {2.1} \right)\left( {2.1} \right)}} = 907,200$
Let us find the total number of words formed using the word INTERMEDIATE if vowels and constraints occupy their original places.
Fix I ,A and E at their given positions. We will apply permutations to the rest of the numbers.
Now, there are 6 letters in which T is repeated 2 times.
Hence, the number of words formed can be calculated as.
$\dfrac{{6!}}{{2!}} = \dfrac{{6.5.4.3.2!}}{{2!}} = 360$
Let us find the total number of words formed using the word INTERMEDIATE if vowels and consonants occur alternatively
There are 6 constants from which T is repeated 2 times and there are 6 vowels from which I is repeated 2 times and E is repeated 3 times.
Also, there can be two possible arrangements, that is when vowels occupy odd places and when vowels occupy even places. So, we multiply the final answer by 2
$\dfrac{{6!}}{{3!2!}} \times \dfrac{{6!}}{{2!}} = 21600$
Let us find the total number of words formed using the word if all the vowels occur together.
Group all the vowels, that is, consider all vowels as a single letter.
Now, we can say that there are 7 letters from which T is repeated 2 times.
Also, the vowels in the group can change their positions and still be together.
Thus, we can say that the number of words formed are,
$\dfrac{{6!}}{{3!2!}} \times \dfrac{{7!}}{{2!}} = 151200$
Hence, option A is correct.

Note: The letters which are fixed should not be used while calculating permutations. Many students forget to take the possibility of both odd and even places the vowels can take. Also, when all the vowels are considered as a single letter, it is important that we find the number of arrangements that vowels can make among themselves.