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 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=m1
    Now, 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

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