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by ivan.m » Thu Jun 16, 2011 4:58 pm
Using the letters of the alphabet, how many three letters codes are possible if a consonant must be the middle letter and different vowels must be used first and last?

1) 2600
2) 2100
3) 525
4) 520
5) 420

The ans: 420
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by winniethepooh » Thu Jun 16, 2011 5:49 pm
which alphabet?
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by ivan.m » Thu Jun 16, 2011 5:50 pm
English one
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by ivan.m » Thu Jun 16, 2011 5:53 pm
Quick start:

I just thought, if order does matter and can be repeated:

n^r => 26^3 this gives all the three codes that can be formed with the english alphabet, but I don't know how to include the restrictions.
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by smackmartine » Thu Jun 16, 2011 6:00 pm
IMO E

Total English alphabets = 26
vowels = 5 (a,e,i,o,u)
Consonants = 26-5 = 21

First position can be filled in in 5 ways
Middle position should be a consonant ,so we have 21 ways, and
Last position can be filled by 4 vowels only other than what we place in 1st position, because question says "different vowels" . So,

# of ways three letters codes are possible = 5 * 21 * 4 = 420
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by ivan.m » Thu Jun 16, 2011 6:11 pm
thanks a lot I was (don't know why :S) substracting the consonant,so I had:

26-5(vowels)= 21-1(consonant)= 20 letters remaining...
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by amit2k9 » Thu Jun 16, 2011 7:08 pm
let the letters be abc.

for a and c = 5*4 = 20
for b = 26-5 = 21

total = 21*20 = 420.
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by edvhou812 » Thu Jun 16, 2011 11:27 pm
26 letters are in the alphabet, and there are 5 vowels A,E,I,O,U, so there are 21 consonants.

First vowel: 5 possibilities
Consonant: 21 possible
Second vowel: 4 possible

5*21*4=420.
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