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# A fairly easy one.. 1. In how many ways can the word

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A fairly easy one.. 1. In how many ways can the word [#permalink]

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22 Jul 2003, 17:00
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This topic is locked. If you want to discuss this question please re-post it in the respective forum.

A fairly easy one..

1. In how many ways can the word CONSTANTINOPLE be arranged?
2. In how many ways can the above word be arranged with vowels and consonants taken together?
3. In how many ways can the word be arranged such that no two vowels are together?
4. Also find probability of a word formed with NO two vowels together?

Last edited by evensflow on 24 Jul 2003, 16:25, edited 2 times in total.
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22 Jul 2003, 22:02
A fairly easy one..

1. In how many ways can the word CONSTANTINOPLE be arranged?
2. In how many ways can the above word be arranged with vowels and consonants taken together?
3. In how many ways can the word be arranged such that no two vowels are together?
4. Also find probability of having NO two consonants together?

the answere for the first one is :

14! / (3!*2!*2!)

Ans 2:

if i interpret u question that consonants taken "appearing togeather " and vowels "appearing togeather"..then we have 5 vowels out of which 2 are same therefore number of ways we can arrange these togeather is ( 5!/2!)

take this as one unit and then arrange it with the consonants which are 9 with 3 and 2 being the same....

so total number of ways u can arrange both sets is

10! / 3!2! multiplied by 5!/2!

10! 5! / 3! 2! 2!..

Ans 3 :

No two vowels are togeather means ..... there are 10 spaces between the 9 consonants...

to make sure that no two vowels are togeather means that vowels needs to be arranged within these 10 spaces ....
10*9*8*7*6 / 2!

and rest of the 9 consonants can be arranged separately 9!/3!2!

10C5 *9! / 3!2!2!

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23 Jul 2003, 16:23
You are pratially right..

I shall post the solution...

But i want few more tries on this problem..
Manager
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24 Jul 2003, 16:25
Okies here's the solution:

The word CONSTANTINOPLE has 9 consonants of which N repeates 3 times, T repeats 2 times and 5 vowels of which O repeates 2 times.

1. 14 letters can be arranged 14! ways and there are some characters repeating so their arramgments are double counted.

So Ans = 14! / 3! * 2! *2!.

2. Vowels and consonants are to be considered to seperate groups. So both can be aranged in 2! ways.

Also consonants can be arranged in 9! / 3! * 2! ways and vowels can be arranged in 5!/ 2! ways.

So Ans = 2 * 9! * 5! / 3! * 2! * 2!

3. if no vowels are together than we start how many places are there for vowels to be there.

There are 9 consonants which can be arranged amongst themselves in 9!/ 3! * 2! ways.

So between the consonants there is places for vowels and there are 10 such places.

Consider this.

_C_C_C_C_C_C_C_C_C_

So, in 10 places vowels can be arranged in 10P5 ways that is 10!/5!. Also there are 2 vowels which are repeated so final ways is 10! / 5! * 2!

So Ans = 10! * 9! / 5! * 3! * 2! *2!

4. Probability of getting a word with NO two vowels is then

Ans 3 / Ans 1.

So Ans = 18/143.
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24 Jul 2003, 16:27
Well, i agree such problems have not been asked in GMAT at all but one has to be prepared.

Also there are high chances that part 4 of the question can appear on GMAT so you never know.
24 Jul 2003, 16:27
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