We are to form different words with the letters of the word INTEGER. Let $m_{1}$ be the number of words in which I and N are never together and $m_{2}$ be the number of words which begin with I and end with R, then $m_{1}/m_{2}$ is equal to

Question

Dr. Abubacker · Accepted Answer

(a)
In the word INTEGER, we have 5 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, ‘I’ and ‘N’ can be placed in 6 available gaps in ${\ ^{6}P}_{2}$ ways.
So, required number of ways $= \frac{5!}{2!}.{\ ^{6}P}_{2} = m_{1}$. Now, if word start with ‘I’ and end with ‘R’, then the remaining letters are 5
So, total number of ways $= \frac{5!}{2!} = m_{2}$
$	herefore\ \frac{m_{1}}{m_{2}} = \frac{5!}{2!}.\frac{6!}{4!}.\frac{2!}{5!} = 30$

Q. We are to form different words with the letters of the word INTEGER. Let $m_{1}$ be the number of words in which I and N are never together and $m_{2}$ be the number of words which begin with I and end with R, then $m_{1}/m_{2}$ is equal to

Explanation

Reviewed by Dr. Abubacker

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