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26 Apr 2017, 08:40
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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:
95%
(hard)
Question Stats:
36% (02:46) correct
64%
(02:45)
wrong
based on 186
sessions
History
Date
Time
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Not Attempted Yet
Using the standard 26-letter alphabet, how many different groups of 4 letters are possible, if exactly two letters are chosen from a, e, i, o ,u and two letters are chosen from all the remaing letters except for y, and a letter can be used more than once?
A. 500
B. 1900
C. 2500
D. 3150
E. 6300
A. 500
B. 1900
C. 2500
D. 3150
E. 6300
02 May 2017, 16:55
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Bunuel
Since the order doesn’t matter and the letter can be repeated, the number of ways to choose 2 vowels is 5C2 + 5C1 = 10 + 5 = 15. (Notice that 5C2 is the number of ways to choose 2 vowels if they are different and 5C1 is the number of ways to choose 2 vowels if they are the same.)
Similarly, the number of ways to choose 2 non-y consonants is 20C2 + 20C1 = 190 + 20 = 210.
Thus, the number of ways to choose 2 vowels and 2 non-y consonants is 15 x 210 = 3150.
Answer: D
General Discussion
it's merely a selection problem:
no. of different ways of selecting 4 letters as per question would be (5c2+5)*(20c2+20)=15*210=3150-> option D ; I have consider the repetitive part and the non-repetitive part of each type i.e vowels and consonant type and the total no. of selections is the product of the no. of selections of these two types.
no. of different ways of selecting 4 letters as per question would be (5c2+5)*(20c2+20)=15*210=3150-> option D ; I have consider the repetitive part and the non-repetitive part of each type i.e vowels and consonant type and the total no. of selections is the product of the no. of selections of these two types.
