First slide

Permutations

Question

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

Moderate

A

30
B

60
C

90
D

180

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 $^{6} P_{2}$ , ways.
So, required number = $= \frac{5!}{2!} \cdot^{6} P_{2} = m_{1}$
Now, if word 8ta!t with 'I' and end with' R, then the remaining letters are 5.
So, total number of $= \frac{5!}{2!} = m_{2}$
$∴ \frac{m_{1}}{m_{2}} = \frac{5!}{2!} \cdot \frac{6!}{4!} \cdot \frac{2!}{5!} = 30$

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