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how can we knw that there will be half times A will cum after Q and Q after A???

m confused ..!!
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how can we knw that there will be half times A will cum after Q and Q after A???

m confused ..!!

Ask yourself: why should any of these two orderings have preferences over the other. Does probability favor one ordering over another?

Follow the links in my previous posts to practice more questions about this concept.

Hope it helps.
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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:

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.

Are there any restrictions while using this approach? Can you please explain?
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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:

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.
Are there any restrictions while using this approach? Can you please explain?

Hi,

The only restriction that you should keep in mind is that how the arrangements are made. In this question, sequence of two alphabets is mentioned. Further extrapolation can be made for three alphabets. For E comes after Q, and Q comes after U, required cases would be 1/6 of total arrangements.
E,Q and U can be arranged in 3! or 6 ways. While the sequence EQU in that order will occur only once.

Hope it helps.

Regards,
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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:

I will never forget the lesson on combinatorics presented by mikemcgarry on Magoosh.
so..we have 8 different letters. these can be arranged in 8! ways.
for every possible way when A comes after Q, there is a combination when A comes before Q. so we need to divide by 2.

8!/2 = 8*7!/2 => 4*7!
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