Question

We are required to form different words with the help of 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 with $R$, then prove that $m_1/m_2=30$

Answer 1

Total number of words =$\frac{7!}{2!}$

Total number of words in which I and N never together$=\frac{6!}{2!}\times 2!$

$m_1=$ Total $-$ together

$=\frac{7!}{2!}-6!=\frac{5}{2}\times 6!$

$m_2=numberofword=\frac{5!}{2!}$
$\therefore \frac{m_1}{m_2}=\frac{5}{2}\times 6!\times \frac{2!}{5!}=6\times 5=30$.