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

# 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

1. A

30

2. B

60

3. C

90

4. D

180

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### 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 , 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 $=\frac{5!}{2!}={\mathrm{m}}_{2}$

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