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04 May 2011, 02:12
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
57% (01:47) correct
43%
(01:38)
wrong
based on 367
sessions
History
Date
Time
Result
Not Attempted Yet
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
A. 210
B. 1050
C. 25200
D. 21400
E. 42800
Guys I've some concerns on this problem and I've put them under the spoiler. Please help me out!Thanks..
A. 210
B. 1050
C. 25200
D. 21400
E. 42800
Guys I've some concerns on this problem and I've put them under the spoiler. Please help me out!Thanks..
I am getting A. However, the OA is C. Below is the OE:
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2)
={(7 x 6 x 5)x(4 x 3)}/ {(3 x 2 x 1)x(2 x 1)}
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.
I don't understand what is need of arranging the 5 letters again, as 7C3*4C2 will do that already. Can someone tell me where I'm going wrong?
Explanation:
Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4) = (7C3 x 4C2)
={(7 x 6 x 5)x(4 x 3)}/ {(3 x 2 x 1)x(2 x 1)}
= 210.
Number of groups, each having 3 consonants and 2 vowels = 210.
Each group contains 5 letters.
Number of ways of arranging
5 letters among themselves = 5!
= 5 x 4 x 3 x 2 x 1
= 120.
Required number of ways = (210 x 120) = 25200.
I don't understand what is need of arranging the 5 letters again, as 7C3*4C2 will do that already. Can someone tell me where I'm going wrong?
22 Apr 2018, 06:38
Kudos
Bookmarks
gmatpapa
Take the task of creating 5-letter words and break it into stages.
Stage 1: Select the 3 consonants to work with
Since the order in which we select the consonants does not matter, we can use combinations.
We can select 3 consonants from 7 consonants in 7C3 ways (= 35 ways)
Stage 2: Select the 2 vowels to work with
Since the order in which we select the vowels does not matter, we can use combinations.
We can select 2 vowels from 4 vowels in 4C2 ways (= 6 ways)
If anyone is interested, we have a video on calculating combinations (like 4C2) in your head - see below
Stage 3: Take the 5 selected letters and arrange them.
We can complete this stage in 5! ways (= 120 ways).
By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create all 5-letter words) in (35)(6)(120) ways (= 25,200 ways)
Answer: C
Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.
RELATED VIDEOS
General Discussion
04 May 2011, 07:40
Kudos
Bookmarks
Using 4 x 3 / (2 x 1) = 6, to select your vowels... just calculated the number of groupings.
Example:
a e i o (4 vowels)
6 POSSIBLE GROUPINGS
a e
a i
a o
e i
e o
i o
If you take the ARRANGEMENT into account,
you should have 12 instead of 6.
a e
a i
a o
e i
e o
i o
e a
i a
o a
i e
o e
o i
But it's best to just get the possible number of grouping first which is 7C3 4C2 = 210. Then we arrange it by multiplying to 5!.. So as to allow consonants and vowels alternating...
Ex.
c d f e i
This will allow c d e i f.. Alternating elements...
Example:
a e i o (4 vowels)
6 POSSIBLE GROUPINGS
a e
a i
a o
e i
e o
i o
If you take the ARRANGEMENT into account,
you should have 12 instead of 6.
a e
a i
a o
e i
e o
i o
e a
i a
o a
i e
o e
o i
But it's best to just get the possible number of grouping first which is 7C3 4C2 = 210. Then we arrange it by multiplying to 5!.. So as to allow consonants and vowels alternating...
Ex.
c d f e i
This will allow c d e i f.. Alternating elements...
