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Find the probability of at least two vowels together when letters in word "AEINCB" are rearranged for all random permutations.

What will be new probabilities when word is changed to either "AEINCBB" or "AEIINCB"?

My Approach so far:

For the word AEINCB, I have tried to sum following arrangements: Two vowels together + 3 vowels together. I can take any of 2 vowels out of 3 and arrange it in 3P2 ways multiplied by remaining letters which can be arranged in 5! ways (4 letters + one vowel group). Similarly followed same approach for 3 vowels together. My answer is 3P2*5! + 3!*4! (3 remaining letters and 1 vowel group). But I feel there is an overlap between these two arrangements and not sure how to remove them. I have not been able to go beyond this for now ("AEINCBB" or "AEIINCB").

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  • $\begingroup$ For the word AEINCB, I have tried following approach. $\endgroup$ – Chandan Jul 12 '15 at 6:31
  • $\begingroup$ I don't see your approach. Do you mean you have solved/tried the first part? If so, show it in the question itself. $\endgroup$ – true blue anil Jul 12 '15 at 6:42
  • $\begingroup$ The AEINCBB is interesting, since I might disagree with the expected answer. $\endgroup$ – André Nicolas Jul 12 '15 at 6:44
  • $\begingroup$ For the word AEINCB, I have tried to sum following arrangements: Two vowels together + 3 vowels together. I can take any of 2 vowels out of 3 and arrange it in 3P2 ways multiplied by remaining letters which can be arranged in 5! ways (4 letters + one vowel group). Similarly followed same approach for 3 vowels together. My answer is 3P2*5! + 3!*4! (3 remaining letters and 1 vowel group). But I feel there is an overlap between these two arrangements and not sure how to remove them. I have not been able to go beyond this for now ("AEINCBB" or "AEIINCB"). Thanks for promptly responding. $\endgroup$ – Chandan Jul 12 '15 at 6:46
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All of your questions can be better handled by computing

1 - P(all the vowels are separate)

AEINCB:

To be separate, the vowels can only be placed in the gaps between consonants (dashes):

- N - C - B -

Thus $^4P_3$ ways to place the vowels, 3! ways to permute consonants in their places,

and Pr = $1 - \dfrac{4\cdot3\cdot2\cdot 3!}{6!} = \dfrac{4}{5}$

AEINCBB

Using the same technique, we have - N - C - B - B -

To be separate, the vowels can be placed in $^5P_3 = 60$ ways, and the consonants permuted in 4!/2!= 12 ways, against a total of 7!/2!=2520 ways,

thus Pr = $1 - \dfrac{60\cdot12}{2520}= \dfrac{5}{7}$

The last one I leave for you !

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  • $\begingroup$ Thanks Anil. Would you validate approach from @barak? $\endgroup$ – Chandan Jul 12 '15 at 7:41

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