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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
(hard)
Question Stats:
41% (02:00) correct
59%
(01:59)
wrong
based on 46
sessions
History
Date
Time
Result
Not Attempted Yet
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) \(_8P_4+\frac{4!}{2!2!}+2*7C2*4!/2!\)
E) \(\frac{_{10}P_2}{2!2!}\)
Could you confirm that my answer is indeed correct ? Many thx in advance.
A) \(_8P_2\)
B) \(\frac{_8P_2}{2!2!}\)
C) \(_8P_4\)
D) \(_8P_4+\frac{4!}{2!2!}+2*7C2*4!/2!\)
E) \(\frac{_{10}P_2}{2!2!}\)
Could you confirm that my answer is indeed correct ? Many thx in advance.
Archived Topic
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28 Feb 2016, 07:01
Kudos
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leeto
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..
26 Jun 2017, 06:42
Kudos
Bookmarks
leeto
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!
