Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2

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.

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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...

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.
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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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7c3*4c2*5!

= 25200

Answer is C.

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.

I think the question needs to mention that repetition of letters is not allowed

Posted from my mobile device
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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

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.

Cheers,
Brent

Given the OA, the prompt should read as follows:



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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