Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
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Question Stats: Hide Show timer StatisticsOut of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?A. 210 B. 1050C. 25200D. 21400E. 42800Guys I've some concerns on this problem and I've put them under the spoiler. Please help me out!Thanks.. _________________
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
gmatpapa wrote: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?A. 210 B. 1050C. 25200D. 21400E. 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: CNote: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it. RELATED VIDEOS_________________
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 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 GROUPINGSa ea ia oe ie oi o If you take the ARRANGEMENT into account, you should have 12 instead of 6.a ea ia oe ie oi oe ai ao ai eo eo iBut 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...
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 7C3 * 4C2 * 5!(7 * 6 * 5)/3! * 4!/2!2! * 12035 * 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.
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
subhashghosh wrote: 7C3 * 4C2 * 5!(7 * 6 * 5)/3! * 4!/2!2! * 12035 * 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! _________________
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 7c3*4c2*5!= 25200 Answer is C.
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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 5 letters in the word - hence 5! ways of arranging them. positions - - - - - 7* 6 *5 * 4 * 3 consonants vowels Thus without actually multiplying one can deduce like this : 5! = 120 means the number is divisible by 3.D and E POE. Also, the number must have 2 zeros at the units and tens place, 120 and 6*5 = 30 is there. C is a perfect fit.
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] I think the question needs to mention that repetition of letters is not allowed Posted from my mobile device _________________
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
firas92 wrote: I think the question needs to mention that repetition of letters is not allowed Posted from my mobile device Agreed! Otherwise, words like AABBB would be allowed. _________________
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Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
GMATPrepNow wrote: firas92 wrote: I think the question needs to mention that repetition of letters is not allowed Posted from my mobile device Agreed! Otherwise, words like AABBB would be allowed. أHi GMATPrepNow Brent How will be the solution if repetition would allowed? what will change in way to solve it?Thanks in advance
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink]
Mo2men wrote: GMATPrepNow wrote: firas92 wrote: I think the question needs to mention that repetition of letters is not allowed Posted from my mobile device Agreed! Otherwise, words like AABBB would be allowed. أHi GMATPrepNow Brent How will be the solution if repetition would allowed? what will change in way to solve it?Thanks in advance
Brent Hanneson – Creator of gmatprepnow.comI’ve spent the last 20 years helping students overcome their difficulties with GMAT math, and the biggest thing I’ve learned is… Many students fail to maximize their quant score NOT because they lack the skills to solve certain questions but because they don’t understand what the GMAT is truly testing - Learn more
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Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 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. 1050C. 25200D. 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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Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] Hello from the GMAT Club BumpBot!Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email. _________________
Re: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 [#permalink] 20 Dec 2021, 05:47 |
Out of 7 consonants and four vowels, how many words of three consonants and 2 vowels can be formed
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