We are to form different words with the letters of the word INTEGER. Let m₁, be the number of words in which I and N are never together and m₂ be the number of words which begin with I and end with R, then m₁/m₂, is equal to

Solution:

In the word INIEGER, we have S letters other than 'I' and ‘N’ of which two are identical (E's). We can arrange
these letters in$\frac{5!}{2!}$ ways. In any such arrangements, 'l' and 'N can be placed in 6 available gaps in ${ }^6{\mathrm{P}}_2$, ways.
So, required number = $=\frac{5!}{2!}{\cdot }^6{\mathrm{P}}_2={\mathrm{m}}_1$
Now, if word 8ta!t with 'I' and end with' R, then the remaining letters are 5.
So, total number of $=\frac{5!}{2!}={\mathrm{m}}_2$

$\therefore \frac{{\mathrm{m}}_1}{{\mathrm{m}}_2}=\frac{5!}{2!}\cdot \frac{6!}{4!}\cdot \frac{2!}{5!}=30$