Out of 7 Consonants and 4 Vowels, How Many Words of 3 Consonants and 2 Vowels Can be Formed? GMAT Problem Solving

Question: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?

1. 210
2. 1050
3. 25200
4. 21400
5. 42800

“Out of 7 Consonants and 4 Vowels, How Many Words of 3 Consonants and 2 Vowels Can be Formed? GMAT Problem Solving”- is a topic of the GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Official Guide Quantitative Review". To solve GMAT Problem Solving questions a student must have knowledge about a good number of qualitative skills. GMAT Quant section consists of 31 questions in total. The GMAT quant topics in the problem-solving part require calculative mathematical problems that should be solved with proper mathematical knowledge.

Solution and Explanation:

Approach Solution 1:

Explanation: considering the given question, what needs to be found is how many words can 3 consonants and 2 vowels out of 7 consonants and 4 vowels can be created. This implies that using 3 consonants and 2 vowels, 5-letter words are supposed to be created. Accordingly, how many words can be created by 5 letters needs to be evaluated focusing on 3 different stages.

In the first stage, it is needed to select the 3 consonants to work with
Since the order in which to select the consonants does not matter, we can use combinations.
We can select 3 consonants from 7 consonants in 7C3 ways (= 35 ways)

The second stage states about selecting the 2 vowels to work with
Since the order in which the selection of the vowels does not matter, we can use combinations.
We can select 2 vowels from 4 vowels in 4C2 ways (= 6 ways)

From here moving on to stage three, 5 selected letters can be taken and arranged.
This stage can be hence completed in 5 ways which imply 120 ways (= 120 ways).

With the need to find the number of words from 3 consonants and 2 vowels, the use of the Fundamental Counting Principle (FCP) can be ensured. Accordingly, by completing all the 3 stages the five letter words can be created in the following number of ways. It depicts the total number of ways that can be multiplied. This implies-

(35)(6)(120) ways = 25,200 way

Hence, option C with 25,200 ways is the correct answer.

Correct Answer: C

Approach Solution 2:

Explanation: With the given case of choosing 3 consonants out of 7 and two vowels out of 4, the number of words that can be formed would include five letters. Accordingly, the positions or arrangement of the letters is not mentioned which means, it can be in any way. For this, the following evaluations need to be considered to find the total number of words.

For the five-letter words with 3 consonants, the number of ways where these can be used in terms of 3 positions can be equated with the combination of 5C3. This can be equated as-

\(\frac{5*4*3}{3*2*1}=10\)

For the remaining 2 positions in the 5 letter words, where the 2 positions would be used by the 2 vowels, the combination that can be identified is 2C2. This can be equated as-

\(\frac{2*1}{2*1}=1\)

Number of options for the first consonant = 7, which can be any of the 7 consonants
Number of options for the second consonant = 6, which can any of the 6 remaining consonants
Number of options for the third consonant = 5 which can be any of the 5 remaining consonants
Number of options for the first vowel = 4 which implies for any of the 4 vowels
Number of options for the second vowel = 3 which implies for any of the 3 remaining vowels

By combining all of these options that have been found from above, all of these values need to be multiplied. This would stand as-

10*1*7*6*5*4*3 = 25,200

Hence, the number of words that can be formed out of 3 consonants and 2 vowels is 25,200 which is the right option for C.

Correct Answer: C

Approach Solution 3:

There are 5 letters in the word

Hence 5! ways of arranging them.

So we get the positions - - - - -
7* 6 *5 * 4 * 3
consonants vowels

We can multiply this or do it by the below process

5! = 120 means the number is divisible by 3.

D and E are properties of equivalence.

Also, the number must have 2 zeros at the units and tens place, 120 and 6*5 = 30 is there.

Hence, C is a perfect fit.

Correct Answer: C

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