How many 5-letter words can be formed using the letters of the English

Question

How many 5-letter words can be formed using the letters of the English alphabet that contain 2 different vowels and 3 different consonants?

A. 5C2 * 21C3
B. 5P2 * 21P3
C. 5C2 * 21C3 * 5!
D. 5P2 * 21P3 * 5!
E. 5^2 * 21^3 * 5!

Kudos for a correct  solution .

KarishmaB · Accepted Answer

No, this is not correct. You are using basic counting principle for selection. Basic counting principle selects AND arranges. For selection, you must use the combination formula 5C2 and 21C3 and then arrange the 5 letters you get as 5!.

When I say that one spot can be filled in 5 ways, I have allotted a spot already. But there is no defined spot for vowels and consonants.

Hence, it is advisable that when you have to select, use combinations formula and then arrange. I rarely use the permutations formula because often one needs to make multiple selections and then arrange them all together. 
I cannot use 5P2 * 21P3 here because again, I am assuming that I have fixed spots for vowels and fixed for consonants.

To avoid confusion, when you only have to select, use combinations.

When you have to select and arrange, use combinations to select and then arrange.

ENGRTOMBA2018 · Answer

Let the arrangement be V1 V2 C1 C2 C3

where V1 and V2 are the vowels (5 in total)
C1,C2,C3 are the consonants (21 in total)

Thus we can select 2 vowels out of 5 available in 5C2 ways and we can select 3 consonants out of 21 in 21C3 ways.

Additionally, the arrangement mentioned above can be arranged in a further 5! ways.

Thus the total number of arrangements = 5!*5C2*21C3. C is the correct answer.

sharma123 · Answer

I think answer is option D , please explain if im wrong  Engr2012

ENGRTOMBA2018 · Answer

can you show your steps of calculation?

sharma123 · Answer

For two vowels we can fill one the five spaces with five vowels ,and for second one we can use 4 vowels so 5*4.
For three consonants we can use 21 consonants for first letter ,20 for second and 19 for third. i.e.21*20*19. these five letters can be arranged in 5! ways which So total no of ways =5*4*21*20*19*5!. I have used (- - - - -) this method . Engr2012

AdlaT · Answer

5 Vowels A,E,I,O,U and 21 Consonants.

it is given that 2 different vowels and 3 different consonants, so repetition not allowed.

we can choose 2 different vowels in  5C2  ways, 3 different consonants in  21C3  and arrange them in  5!  ways.

Total 5 letter words  5C2*21C3*5! .

Ans.C.

Additional. if the word "different" is not given,

Total 5 letter words  5*5*21*21*21 .

PolarCatastrophe · Answer

Actually things get much messier than that:

How many 5-letter words can be formed using the letters of the English alphabet that contain 2 different vowels and 3 different consonants?

is now

How many 5-letter words can be formed using the letters of the English alphabet that contain 2 vowels and 3 consonants?

What you've suggested in  5*5*21*21*21  is the number of unique values with set positions for vowels and consonants; VVCCC. You might suggest the way to fix this is just 5!. However, when repetition is allowed within the V and the C, you will get duplicates in your words and have to correct for this.

For example, say you get:
AABBB can only form  5!/(2!3!)  = 10 different words.
AEBCD can form 5! = 120 different words.

We have to break it down a bit more: 
Vowels: Since there are 5, there are 5*5 = 25 different two vowel combinations. 5 of these are pairs: AA, EE, etc. The other 20 are both different.
Consonants: Since there are 21, there are 21*21*21 = 9261. 21 of these are triplets: BBB, CCC, DDD and 3*20*21 = 1260 that contain doubles. (This can be seen by @@#, @#@, #@@ where @ can be any of the 21 consonants and # is any of the remaining 20 consonants.) There are 21*20*19 = 7980 where all are different.

Returning to the original setup we need to choose 2 vowels and 3 consonants. They could have any of these forms:
(Vowels) (Consonants)
(Both Different)(All Different) = 20*7980
(Both Different)(Two the Same) = 20*1260
(Both Different)(All the Same) = 20*21
(Pair)(All Different) = 5*7980
(Pair)(Two the Same) = 5*1260
(Pair)(All the Same) = 5*21

And for each of these scenarios there is a different amount of unique words that can be created:
(Both Different)(All Different) = 5!
(Both Different)(Two the Same) = 5!/2!
(Both Different)(All the Same) = 5!/3!
(Pair)(All Different) = 5!/2!
(Pair)(Two the Same) = 5!/(2!2!)
(Pair)(All the Same) = 5!/(2!3!)

So you need to take the two distributions above to get the total number of unique words given repetition in choosing vowels/consonants.

This is why you can almost always assume the GMAT has a pretty basic combinatorics problem, because a slight change like this would make the problem very long. =)

sharma123 · Answer

VeritasPrepKarishma . is my approach correct here??
Also please explain where to use combinations/permutations.

HKHR · Answer

Total vowels = 5
Total consonants = 21

The word is of the form v1 v2 c1 c2 c3. This word can be arranged in  5!  ways (as the vowels and the consonants are different).

The 2 out of 5 vowels can be chosen in  5C2  ways.
3 out of 21 consonants can be chosen in  21C3  ways.

Total number of 5 letter words =  5!  *  5C2  *  21C3

Option C is the right answer

KS15 · Answer

2 different vowels can be selected from 5 in 5c2 ways
3 different consonants can be selected from 21 in 21c3 ways
Also, we will need to arrange the 5 alphabets in 5 ! ways
Answer=5c2*21c3*5!
C

AkshdeepS · Answer

My solution:

As we have 5 vowels and 21 consonants then,

Choosing 2 vowels from a set of 5 = 5C2  ways

Choosing 3 consonants from a set of 21 =  21C5  ways

Now the permutation of 5 letters or we can arrange these 5 letters in 5! ways (Important Step)
 5C2 * 21C3 *5!

Then answer is Option C

Crytiocanalyst · Answer

Chosing the 2 vowels in 5c2 ways

remaining 3 consonants in 21C3 ways

and arranging all of them in 5!

Thereofore IMO C

Kinshook · Answer

Asked:  How many 5-letter words can be formed using the letters of the English alphabet that contain 2 different vowels and 3 different consonants?

Ways to select 2 different vowels out of 5 different vowels =  ^5C_2 
Ways to select 3 different consonants out of 21 different consonants =  ^{21}C_3 
Ways to arrange selected 2 different vowels and 3 different consonants = 5!
The number of 5-letter words can be formed using the letters of the English alphabet that contain 2 different vowels and 3 different consonants =  ^5C_2 * ^{21}C_3 * 5!

IMO C

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How many 5-letter words can be formed using the letters of the English

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