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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 $m_1$ (words where $I$ and $N$ are never toge

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) The word $INTEGER$ contains $7$ letters: $I, N, T, E, G, E, R$. The letter $E$ repeats twice.To find $m_1$ (words where $I$ and $N$ are never together):Total arrangements of $INTEGER$ is $\frac{7!}{2!} = 2520$.Arrangements where $I$ and $N$ are together: Treat $(IN)$ as one unit. We have $6$ units: $(IN), T, E, G, E, R$. These can be arranged in $\frac{6!}{2!} 	imes 2! = 720$ ways.So,$m_1 = 2520 - 720 = 1800$.To find $m_2$ (words starting with $I$ and ending with $R$):Fix $I$ at the first position and $R$ at the last position. The remaining $5$ letters are $N, T, E, G, E$. These can be arranged in $\frac{5!}{2!} = 60$ ways.So,$m_2 = 60$.Therefore,$\frac{m_1}{m_2} = \frac{1800}{60} = 30$.

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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