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How many different four-letter words can be formed (the words need not

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How many different four-letter words can be formed (the words need not  [#permalink]

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New post Updated on: 02 Sep 2018, 09:16
6
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A
B
C
D
E

Difficulty:

  85% (hard)

Question Stats:

42% (01:24) correct 58% (01:27) wrong based on 64 sessions

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How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS ?

A) [m]_8P_2[/m]

B) [m][fraction]_8P_2/2!2![/fraction][/m]

C) [m]_8P_4[/m]

D) [m]_8P_4+[fraction]4!/2!2![/fraction]+2*7C2*4!/2![/m]

E) [m][fraction]_{10}P_2/2!2![/fraction][/m]


Could you confirm that my answer is indeed correct ? Many thx in advance.

Originally posted by leeto on 28 Feb 2016, 06:25.
Last edited by chetan2u on 02 Sep 2018, 09:16, edited 3 times in total.
RENAMED THE TOPIC and edited OA
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Re: How many different four-letter words can be formed (the words need not  [#permalink]

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New post 28 Feb 2016, 07:01
leeto wrote:
How many different four-letter words can be formed (the words need not be meaningful) using
the letters of the word GREGARIOUS ?

A) \(_8P_2\)

B) \(\frac{_8P_2}{2!2!}\)

C) \(_8P_4\)

D) \(\frac{_8P_4}{2!2!}\)

E) \(\frac{_{10}P_2}{2!2!}\)


Could you confirm that my answer is indeed correct ? Many thx in advance.


Hi,
I do not think D can be the answer..
GREGARIOUS has 10 letters, out of which two G and R are used twice..
so if I take only one of G and R, I will have 8 different letters..


I can choose 4 letters out of these 8 in 8C4 ways and these 4 can be arranged in 4! amongst themselves, thus 8C4*4!=8P4 ways..
Now you have other ways where the word can have two same letters and other two different..
so obviously answer will be more than 8P4 but D says 8P4/2!2!, which is 1/4 th of 8P4..
Not possible that D is the answer..

You will get the answer in 3 ways..

1) all letters different- 8P4
2) two of one kind and other two different-- 7C2 * 4!/2! *2
3) two of one kind and two of other-4!/2!2!..

ADD all three to get the answer..

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html


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Re: How many different four-letter words can be formed (the words need not  [#permalink]

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New post 26 Jun 2017, 06:42
leeto wrote:
How many different four-letter words can be formed (the words need not be meaningful) using the letters of the word GREGARIOUS ?

A) \(_8P_2\)

B) \(\frac{_8P_2}{2!2!}\)

C) \(_8P_4\)

D) \(\frac{_8P_4}{2!2!}\)

E) \(\frac{_{10}P_2}{2!2!}\)


Could you confirm that my answer is indeed correct ? Many thx in advance.


Responding to a pm:

2G, 2R, A, E, I, O, U, S

You will need to make cases - the 3 different ways in which you can make 4 letter words:

Case 1: 2 Same, 2 Same - Pick G and R and arrange them as 4!/2!*2!

Case 2: 2 Same, 2 Different - Pick one of G and R in 2 ways. Of the remaining 7 distinct letters, pick any 2 in 7C2 ways. Arrange them in 4!/2! ways

Case 3: All 4 different - Out of 8 distinct letters, pick any 4 in 8C4 ways. Arrange them in 4! ways.

Total number of ways = 4!/2!*2! + 2 * 7C2 * 4!/2! + 8C4 * 4!
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Re: How many different four-letter words can be formed (the words need not  [#permalink]

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New post 02 Sep 2018, 04:49
Waiting for the correct OA and explanation. Thanks in advance :D
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Re: How many different four-letter words can be formed (the words need not &nbs [#permalink] 02 Sep 2018, 04:49
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