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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?
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2
[#permalink]
04 May 2011, 06:40
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...
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2
[#permalink]
04 May 2011, 07:16
subhashghosh wrote:
7C3 * 4C2 * 5!
(7 * 6 * 5)/3! * 4!/2!2! * 120
35 * 6 * 120
= 210 * 120
= 25200
I don't understand what is need of arranging the 5 letters again, as 7C3*4C2 will do that already.
7C3*4C2 will select letters, thereafter one has to arrange those.
Uh.. thanks I think I got it now..
Say for example, the first combination is "r t y u i" You can sure arrange it in 5! ways, and since this is a unique combination, alphabets can be arranged in 5! and still form words not contained in any of the other combinations. Thanks for pointing out. great tip! _________________
My GMAT debrief: http://gmatclub.com/forum/from-620-to-710-my-gmat-journey-114437.html
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2
[#permalink]
22 Apr 2018, 05:38
1
Kudos
Expert Reply
Top Contributor
gmatpapa wrote:
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
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.
How will be the solution if repetition would allowed? what will change in way to solve it?
Thanks in advance
That rewording would require us to consider many different cases: 5-letter words with 2 different vowels and 3 different consonants 5-letter words with 2 identical vowels and 3 different consonants 5-letter words with 2 different vowels, 2 identical consonants, and 1 different consonant 5-letter words with 2 identical vowels, 2 identical consonants, and 1 different consonant 5-letter words with 2 different vowels, and 3 identical consonants 5-letter words with 2 identical vowels, and 3 identical consonants
Phew! I'm already tired. As you might imagine, the reworded question would take up wayyyyyyy too much time to be a true GMAT question.
Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2
[#permalink]
31 Oct 2020, 07:31
Expert Reply
Given the OA, the prompt should read as follows:
gmatpapa wrote:
Out of 7 DISTINCT consonants and 4 DISTINCT vowels, how many words of 3 consonants and 2 vowels can be formed, if no letter may appear in the word more than once?
A. 210 B. 1050 C. 25200 D. 21400 E. 42800
One approach:
From the 5 positions in the word, the number of ways to choose 3 positions for the 3 consonants = 5C3 = \(\frac{5*4*3}{3*2*1} = 10\) From the remaining 2 positions in the word, the number of ways to choose 2 positions for the 2 vowels = 2C2 \(= \frac{2*1}{2*1} = 1\)
Number of options for the first consonant = 7 (Any of the 7 consonants) Number of options for the second consonant = 6 (Any of the 6 remaining consonants) Number of options for the third consonant = 5 (Any of the 5 remaining consonants) Number of options for the first vowel = 4 (Any of the 4 vowels) Number of options for the second vowel = 3 (Any of the 3 remaining vowels)
To combine the options above, we multiply: 10*1*7*6*5*4*3 = 25200
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