Permutation of word "UNBIASED" with condition

• Feb 17th 2011, 04:52 AM
vivek_master146
Permutation of word "UNBIASED" with condition
Consider the word UNBIASED. How many words can be formed with the letters of the
word in which no two vowels are together?

If we arrange consonant first it will be:-

4*3*2*1 = 24

Now we are arrange 4 vowels as:-

X C X C X C X C X

Permutation of 5 objects taken 4 at a time. 5! = 120

Answer is coming to be 120 * 24 = 2880

But the actual answer is different.
• Feb 17th 2011, 05:18 AM
What have you mentioned as X here?

this is my attempt check whether this is correct

If V is a vowel, and N is a non-vowel, the arrangment should be,

V,N,V,N,V,N,V,N or N,V,N,V,N,V,N,V

so for vowels no of arrangments - 4!
for non-vowels no of arrangments - 4!

total no. of arrangments = (4!.4!)2
• Feb 17th 2011, 06:03 AM
hirenbhatt
Not very sure but let me try.

The total number of permutations of the letters in the word "UNBIASED" are 8!

1. Now consider first event: take all the vowels together and make it one word
(UIAE) N B S D. These 5 objects can be arranged in 5! ways. But again the 4
vowels can be arranged in 4! ways so total
5! * 4! = 120 * 24 = 2880

2. The event where no two vowels are together, is a mutual exclusive event to the
1st one. So total number of arrangements where no two vowels are together
8! - (5!*4!) = 40320 - 2880 =37440

Is that the answer you are looking for?
• Feb 17th 2011, 06:19 AM
I think you've only excluded the arrangments where all vowels are together, so what about arrangments like UIBNASED and UIEBNASD
• Feb 17th 2011, 07:37 AM
Plato
Quote:

Originally Posted by vivek_master146
Consider the word UNBIASED. How many words can be formed with the letters of the
word in which no two vowels are together?
Answer is coming to be 120 * 24 = 2880
But the actual answer is different.

The four consonants create five places to put the four vowels.
$\displaystyle \underline{\quad}C\underline{\quad}C\underline{\qu ad}C\underline{\quad}C\underline{\quad}$
We can have two consonants together.
So we get $\displaystyle 5\cdot(4!)^2=2880$

Why do you say the actual answer is different from that?
• Feb 17th 2011, 08:41 AM
vivek_master146
Thanks plato. I also computed it in the same say. The answer given in the book is 1152. So i was confused! R u sure that this is the correct answer. ?
• Feb 17th 2011, 09:00 AM
Plato
Quote:

Originally Posted by vivek_master146
The answer given in the book is 1152. So i was confused! R u sure that this is the correct answer. ?

I suspect that whoever worked out that answer assumed that even though it is not correct. Because it does not count $\displaystyle UNBASIDE$ for example.