Question

We are required to form different words with the help of the letters of the word INTEGER. Let ${m_1}$ be the number of words in which I and N are never together and ${m_2}$ be the number of words which begin with I and end with R, then $\dfrac{{{m_1}}}{{{m_2}}}$ is given by:
(a) $30$
(b) $60$
(c) $90$
(d) $120$

Answer 1

Hint: Before we proceed into the problem, it is important to know the definitions of combinations because we will be using this concept.Combination is a mathematical technique that determines the number of possible arrangements where the order of the selection does not matter. Apply the formula of combination and try to find the ratio accordingly.

Formula Used: C= ${}^n{C_r} \times 2!$

Complete step by step Solution:
In the word INTEGER, there are two letters that come more than one time. In this, the letter ‘E’ comes twice. Also, we will use the concept that the no. of words in which I and N are never together is equal to the Total no. of words formed minus when I and N are together.
Given, the word INTEGER
Total letters$ = 7$
First, we will calculate the total number of words that can be formed using INTEGER
We have a combination formula that, if we are taking r letters from n letters in which 2 letters are repeating, then it can be done in ${}^n{C_r} \times 2!$ ways i.e., $ = \dfrac{{7!}}{{2!}}$
$ \Rightarrow \dfrac{{(6! \times 5)}}{{2!}}$
A number of words in which I and N are never together = Total words – I & N are together.
$ \Rightarrow {m_1} = \dfrac{{7!}}{{2!}} - (\dfrac{{6! \times 2}}{{2!}})$
The number of words begin with I and end with R
$ \Rightarrow {m_2} = \dfrac{{5!}}{{2!}}$
Therefore, the ratio $\dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{\dfrac{{(6! \times 5)}}{{2!}}}}{{\dfrac{{5!}}{{2!}}}} = 30$

Hence, the correct option is A.

Note: In these types of questions, we should always use combination formulas, and keep in mind the repeated letters. Also, about the concept of rejection. If we would have solved for I and N never together it can create too much calculations and thought process. Combinations cannot be confused with permutations. In permutations, the order of the selected items is essential.