Total number of ways of arranging the word ALGEBRA is \(\frac{7!}{2!}\)

Now the consonants occupy the same position as in the word ALGEBRA. So consonants will be _LG_BR_.

The 3 vowels in the 3 blanks can be arranged in \(\frac{3!}{2!}\) ways.

The consonants can be arranged in those positions in 4!

Probability = \(\frac{(\frac{3!}{2!}) * 4!}{\frac{7!}{2!}}\)

= \(\frac{3*4!}{3*4*5*6*7}\)

= \(\frac{1}{35}\)

OPTION: E

ALGEBRA ;
total Probability ; 7!/2!
Consonants positions can be taken in 4! ways
and vowels 3!/2! ways arrangement
so
P ; 4!*3!/2! / 7!/2!
solve we get
4!*3/7! ; 1/35
IMO E

Archit3110 Why did you divide total probability by 2?

We have fewer vowels, so it's slightly easier to answer the equivalent question "what is the probability the vowels stay in the same place". If we choose where to put the vowels in our new word, we have seven positions to choose from (first letter, second letter, etc) and we're choosing three of those places. We can choose three things from seven in 7C3 = (7)(6)(5)/3! = 35 ways. In only one of those ways have the vowels remained in their previous locations, so the probability the vowels are where vowels used to be (and thus that consonants are where consonants used to be) is 1/35.
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Fill in the vowels first then consonants positions

Vowels positions can be filled in 3 ways

there are 4 consonants they can be filled in 4! ways

Multiplying 3 by 24 gives 72 ways to arrange the letters given the restrictions. Total ways to arrange the letters ALGEBRA is 7!/2!

72/(7! /2!) gives 1/35

Probability method of solving:

Question requires the consonants occupying the same positions (they can interchange - a mistake I did at first). And if the consonants are at their positions, then the vowels would also need to be in their respective positions (and they can interchange as well).
There are initially 3 vowels and 4 consonants.

Hence position-wise probability = \(\frac{3}{7}*\frac{4}{6}*\frac{3}{5}*\frac{2}{4}*\frac{2}{3}*\frac{1}{2}*1 = \frac{1}{35}\)

Answer E.
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any other way

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