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Standard XII

Mathematics

Permutation: n Different Things Taken All at a Time When All Are Not Different.

We are requir...

Question

We are required to form different words with the help of 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 given by:

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Solution

The correct option is A 30
The word is INTEGER.
Number of ways in which I and N are together $= \frac{6!}{2!} \times 2!$ $= 720$
m1 $= \frac{7!}{2!} - 720 = 1800$
m2 $= \frac{5!}{2!} = 60$
$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{1800}{60} = 30$

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We are required to form different words with the help of 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 given by:

The correct option is A 30The word is INTEGER.Number of ways in which I and N are together =6!2!×2! =720m1=7!2!−720=1800m2=5!2!=60⇒m1m2=180060=30

The correct option is A 30
The word is INTEGER.
Number of ways in which I and N are together $= \frac{6!}{2!} \times 2!$ $= 720$
m1 $= \frac{7!}{2!} - 720 = 1800$
m2 $= \frac{5!}{2!} = 60$
$\Rightarrow \frac{m_{1}}{m_{2}} = \frac{1800}{60} = 30$