in WONKCUL, such that

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in WONKCUL, such that

by sanju09 » Sat Jul 10, 2010 5:47 am
How many words, with or without meanings, can be constructed using all the letters in WONKCUL, such that there are exactly 3 letters between the 2 vowels?
(A) 5!
(B) 5! × 2!
(C) 6!
(D) 6! × 2!
(E) 6! × 3!
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by kvcpk » Sat Jul 10, 2010 5:56 am
Fix 2 places for vowels and then fill the consonants.

So Consonants can be fille din 5! ways.
Vowels can be arranged in 2! ways.

Hence B.

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by sanju09 » Sat Jul 10, 2010 5:58 am
kvcpk wrote:Fix 2 places for vowels and then fill the consonants.

So Consonants can be fille din 5! ways.
Vowels can be arranged in 2! ways.

Hence B.
make sure all conditions were met
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by kvcpk » Sat Jul 10, 2010 6:13 am
sanju09 wrote:
kvcpk wrote:Fix 2 places for vowels and then fill the consonants.

So Consonants can be fille din 5! ways.
Vowels can be arranged in 2! ways.

Hence B.
make sure all conditions were met
My bad!! Answer should be 6!
there are three possible places for the vowels.
So 3*2!*5! = 6!

Right this time?

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by diebeatsthegmat » Sat Jul 10, 2010 11:54 am
sanju09 wrote:How many words, with or without meanings, can be constructed using all the letters in WONKCUL, such that there are exactly 3 letters between the 2 vowels?
(A) 5!
(B) 5! × 2!
(C) 6!
(D) 6! × 2!
(E) 6! × 3!
there are 5! possible ways for letters cos there are 5 letters in the words
there are 3! possible ways for vowels
O---U-- ( 0 is in the first place)
if O is in the second place _O---U-
if O in the third place --O---U
we did the same with U ( just opposite it)
so there are 3!=1*2*3=6 possible ways for those place
thus total we have 6!
answer should be C