We are to form different words with the lette | vedclass

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(A) The word $INTEGER$ contains $7$ letters: $I, N, T, E, G, E, R$. The letter $E$ repeats twice.To find $m_1$ (where $I$ and $N$ are never together),

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$30$

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(A) The word $INTEGER$ contains $7$ letters: $I, N, T, E, G, E, R$. The letter $E$ repeats twice.To find $m_1$ (where $I$ and $N$ are never together),we use the gap method. First,arrange the remaining $5$ letters $(T, E, G, E, R)$. The number of ways to arrange these is $\frac{5!}{2!} = 60$.There are $6$ gaps created by these $5$ letters (including ends). We place $I$ and $N$ in these $6$ gaps in $^6P_2$ ways.$m_1 = \frac{5!}{2!} 	imes ^6P_2 = 60 	imes (6 	imes 5) = 60 	imes 30 = 1800$.To find $m_2$ (words starting with $I$ and ending with $R$),we fix $I$ at the first position and $R$ at the last position. The remaining $5$ letters are $N, T, E, G, E$. The number of ways to arrange these is $\frac{5!}{2!} = 60$.$m_2 = 60$.Therefore,$m_1/m_2 = 1800 / 60 = 30$.

We are to form different words with 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 $m_1/m_2$ is equal to

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