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

In the word INIEGER, we have S letters other than 'I' and ‘N’ of which two are identical (E's). We can arrangethese letters in5!2! ways. In any such arrangements, 'l' and 'N can be placed in 6 available gaps in  6P2, ways.So, required number = =5!2!⋅6P2=m1Now, if word 8ta!t with 'I' and end with' R, then the remaining letters are 5.So, total number of =5!2!=m2∴ m1m2=5!2!⋅6!4!⋅2!5!=30