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16 Jun 2012, 20:38
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
62% (01:33) correct
38%
(01:53)
wrong
based on 410
sessions
History
Date
Time
Result
Not Attempted Yet
How many words can be formed using all the letters of "EQUATION" such that A always comes after Q (not necessarily together)?
A. 7!
B. 2*7!
C. 4*7!
D. 8!/3
E. 8!/6
This question is created by me, I couldn't find similar question in the forum:
A. 7!
B. 2*7!
C. 4*7!
D. 8!/3
E. 8!/6
This question is created by me, I couldn't find similar question in the forum:
17 Jun 2012, 02:21
Kudos
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cyberjadugar
How many words can be formed using all the letters of "EQUATION" such that A always comes after Q (not necessarily together)?
A. 7!
B. 2*7!
C. 4*7!
D. 8!/3
E. 8!/6
This question is created by me, I couldn't find similar question in the forum:
A. 7!
B. 2*7!
C. 4*7!
D. 8!/3
E. 8!/6
This question is created by me, I couldn't find similar question in the forum:
Responding to a pm.
Since there are 8 distinct letters in the word EQUATION, then the # of arrangements of these letters would be 8!. Now, in half of these cases A will be after Q and in half of these cases Q will be after A.
Therefore the final answer is \(\frac{8!}{2}=\frac{7!*8}{2}=4*7!\).
Answer: C.
Questions about the same concept to practice:
six-mobsters-have-arrived-at-the-theater-for-the-premiere-of-the-126151.html
susan-john-daisy-tim-matt-and-kim-need-to-be-seated-in-130743.html
meg-and-bob-are-among-the-5-participants-in-a-cycling-race-58095.html
in-how-many-different-ways-can-the-letters-a-a-b-91460.html
mary-and-joe-are-to-throw-three-dice-each-the-score-is-the-126407.html
goldenrod-and-no-hope-are-in-a-horse-race-with-6-contestants-82214.html
Hope it helps.
General Discussion
16 Jun 2012, 21:38
Kudos
Bookmarks
Answer C.
There are 8 unique alphabets, so 8! ways of arranging them.
For the restriction of A coming after Q -
Among all possible arrangements, in half of the cases A will be before Q - and in the other half it will be after Q.
So answer = 8! / 2 = 8 / 2 x 7! = 4 x 7!
There are 8 unique alphabets, so 8! ways of arranging them.
For the restriction of A coming after Q -
Among all possible arrangements, in half of the cases A will be before Q - and in the other half it will be after Q.
So answer = 8! / 2 = 8 / 2 x 7! = 4 x 7!
