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In how many ways can the letters of the word "COMPUTER" be arranged?
1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vowels occupy the even positions[/b]
1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vowels occupy the even positions[/b]
19 Feb 2012, 04:27
Kudos
Bookmarks
On how many ways can the letters of the word "COMPUTER" be arranged?
1. Without any restrictions:
Since all letters in the word "COMPUTER" are distinct then the # of arrangements is 8!.
2. M must always occur at the third place:
M is fixed at the third place, other 7 distinct letters can be arranged in 7! ways,
3. All the vowels are together:
Consider three vowels as one unit: {OEU}. Thus we'll have total of 6 units: {OEU}{C}{M}{P}{T}{R}, which can be arranged in 6! ways. Three vowels within their unit can be arranged in 3! ways. Total: 6!*3!.
4. All the vowels are never together:
Total minus restriction: 8!-6!*3!.
5. Vowels occupy the even positions (the vowels can occupy only even positions):
C|O|M|P|U|T|E|R
O|E|O|E|O|E|O|E (O and E stand for odd and even positions respectively).
# of arrangements would be \(C^3_4*3!*5!=4!*5!=2880\).
\(C^3_4\) - choosing which 3 even positions out of 4 will be occupied by vowels (there are 4 even positions: 2nd, 4th, 6th and 8th and only 3 vowels);
\(3!\) - # of different arrangements of these vowels on their even positions;
\(5!\) - # of different arrangements of 8-3=5 other letters left.
Hope it helps.
1. Without any restrictions:
Since all letters in the word "COMPUTER" are distinct then the # of arrangements is 8!.
2. M must always occur at the third place:
M is fixed at the third place, other 7 distinct letters can be arranged in 7! ways,
3. All the vowels are together:
Consider three vowels as one unit: {OEU}. Thus we'll have total of 6 units: {OEU}{C}{M}{P}{T}{R}, which can be arranged in 6! ways. Three vowels within their unit can be arranged in 3! ways. Total: 6!*3!.
4. All the vowels are never together:
Total minus restriction: 8!-6!*3!.
5. Vowels occupy the even positions (the vowels can occupy only even positions):
C|O|M|P|U|T|E|R
O|E|O|E|O|E|O|E (O and E stand for odd and even positions respectively).
# of arrangements would be \(C^3_4*3!*5!=4!*5!=2880\).
\(C^3_4\) - choosing which 3 even positions out of 4 will be occupied by vowels (there are 4 even positions: 2nd, 4th, 6th and 8th and only 3 vowels);
\(3!\) - # of different arrangements of these vowels on their even positions;
\(5!\) - # of different arrangements of 8-3=5 other letters left.
Hope it helps.
25 Jun 2007, 19:39
Kudos
Bookmarks
3) All vowels are together -
C-O-M-P-U-T-E-R has 3 vowels - for them to be together, consider them as a single entity say K,
so now we have 6 alphabets (C,M,P,T,R,K) - 6! ways
K comprises of 3 alphates - so K can arrange itself in 3! ways,
hence total 6! x 3!
5) There are 4 even positions to be filled by 3 vowels -so by direct counting 4 x 3 x 2
remaining 5 positions are occupied by the 5 alphabets in 5! ways
=> 5! x 4!
C-O-M-P-U-T-E-R has 3 vowels - for them to be together, consider them as a single entity say K,
so now we have 6 alphabets (C,M,P,T,R,K) - 6! ways
K comprises of 3 alphates - so K can arrange itself in 3! ways,
hence total 6! x 3!
5) There are 4 even positions to be filled by 3 vowels -so by direct counting 4 x 3 x 2
remaining 5 positions are occupied by the 5 alphabets in 5! ways
=> 5! x 4!
) 
