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

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Infinity Learn · Accepted Answer

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 $$=5$$!$$2$$!$$=m2$$∴ $$m1m2=5$$!$$2$$!⋅$$6$$!$$4$$!⋅$$2$$!$$5$$!$$=30$$

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

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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

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Ready to Test Your Skills?

free study material

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