IIT JEE , AIEEE Forum - Mathematics , Algebra

amanks

letters of the word mathematics are arranged in arbitrary fashion then the probability that all the vowels and consonants occupy their previous position with no letter at its previous position?

Decoder

first of all make a box figure with dashes..

M _ T H _ M _ T_C S....

now take vowels first ...
one by one..
A can occupy only too places now..(earlier occupied by i,e)..

since they r alike thus this is only one option..

now ..e have three options ..i also have three ..

so..total no. of options for vowels r 9..

now take consonants the same way..

kaymant

The word is MATHEMATICS.

We have 2 M's, 2 A's, 2 T's, 1 H, 1 E, 1 I, 1 C, 1 S. The vowels are A, E, I with A being repeated twice.

First, the total number of permutations of this word are $\dfrac{11!}{2!\,2!\,2!}$ , which, therefore, becomes the

size of the sample space.

Next, we determine the no. of favorable events. The location of the vowels and consonants should be as follows

_ V _ _ V _ V _ V _ _

with the occurence of V deonting the presence of the vowel there. The four vowels (of which two are A's) can be arranged in the four respective locations in $\dfrac{4!}{2!}$ ways, while the consonants can be arranged in their respective locations in $\dfrac{7!}{2!\,2!}$ . So, the total number of arrangements in which the vowels and consonants occupy their original positions is

$\dfrac{4! \,7!}{2!\,2!\,2!}$

But, out of these permutations, exactly one will be there in which the letters will be in their original position (that is, we shall get the word MATHEMATICS itself). Leaving this out, the required number of favourable ways become

$\dfrac{4! \,7!}{2!\,2!\,2!}-1$

Accordingly, the required probability becomes

$\dfrac{\frac{4! \,7!}{2!\,2!\,2!}-1}{\frac{11!}{2!\,2!\,2!}}=\dfrac{15119}{4989600}\approx 0.003$

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