# We are required to form different words with the help of letters if 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 with R, then prove that $\dfrac{{{m_1}}}{{{m_2}}} = 30$

Last updated date: 27th Mar 2023

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Answer

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Hint: Use permutation to find number of possible words and then proceed further. If two letters are always together count it as one letter and proceed further.

Given word: INTEGER

In this word, number of letters are 7 and ‘E’ is repeated 2 times.

So, all possible arrangements of words are ${}^7{P_2}$.

We know that, ${}^n{P_r} = \dfrac{{n!}}{{r!}}$

\[\therefore {}^7{P_2} = \dfrac{{7!}}{{2!}} = 2520\] $\left( {{\text{using permutation}}} \right)$

Now, For ${m_1}$,

If I and N are always together i.e. “IN” and “NI”, number of possible words are $2.{}^6{P_2} = 2.\dfrac{{6!}}{{2!}} = 720$

Therefore, from given statement, ${m_1}$=$2520 - 720 = 1800$

For ${m_2}$,

If the word is starting with I and ending with R, then ${m_2} = {}^5{P_2} = \dfrac{{5!}}{{2!}} = 60$

$\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{1800}}{{60}} = 30$

Hence Proved.

Note: Whenever you have to find out the number of arrangements from a given set of digits or alphabets, always use permutation. An assortment or a combination of things from a set where the arrangement of the selected things does matter is referred to as permutation. Thus, permutation refers to an ordered combination.

Given word: INTEGER

In this word, number of letters are 7 and ‘E’ is repeated 2 times.

So, all possible arrangements of words are ${}^7{P_2}$.

We know that, ${}^n{P_r} = \dfrac{{n!}}{{r!}}$

\[\therefore {}^7{P_2} = \dfrac{{7!}}{{2!}} = 2520\] $\left( {{\text{using permutation}}} \right)$

Now, For ${m_1}$,

If I and N are always together i.e. “IN” and “NI”, number of possible words are $2.{}^6{P_2} = 2.\dfrac{{6!}}{{2!}} = 720$

Therefore, from given statement, ${m_1}$=$2520 - 720 = 1800$

For ${m_2}$,

If the word is starting with I and ending with R, then ${m_2} = {}^5{P_2} = \dfrac{{5!}}{{2!}} = 60$

$\therefore \dfrac{{{m_1}}}{{{m_2}}} = \dfrac{{1800}}{{60}} = 30$

Hence Proved.

Note: Whenever you have to find out the number of arrangements from a given set of digits or alphabets, always use permutation. An assortment or a combination of things from a set where the arrangement of the selected things does matter is referred to as permutation. Thus, permutation refers to an ordered combination.

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