How many seven-letter words can be constructed using 26 letters of the

Four non vowels can be selected in 21C4 ways. 3 vowels can be selected in 5C3 ways. Since repetition is allowed, therefore the number of ways to arrange these 7 letters can be 7^7.
Therefore the total no. of ways should be 21C4*5C3*7^7.

Experts are requested to provide the solution for this problem.

Not an expert but the highlighted part is wrong since repetition is there. It would be \(21^4\) and \(5^3\) respectively. Had repetition been not there, then it would have been correct.
The third highlighted part is altogether wrong. It is as if you are choosing alphabets from a lot of 7 with repetition, which is not the case.
Also, the 7 letters can be arranged in 7! ways

So, following cases arrive:
1. No repetition is there = \(21_{C_4}*5_{C_3}*7!\)
2. One consonant is repeated but no vowels are repeated = \(21^2*20_{C_3}*5_{C_3}*\frac{7!}{1!}\)
3. Two consonants are repeated but no vowels are repeated = \(21^3*20_{C_2}*5_{C_3}*\frac{7!}{2!}\)
4. Three .....
......
. All the consonants and vowels repeat = \(21^4*5^3*\frac{7!}{4!*3!}\)
...
...
...
X. No consonants are repeated but all vowels are repeated = \(21_{C_4}*5^3*\frac{7!}{3!}\)

The sum of all of these cases would give us the required answered which is not given.

Hence E is the answer.

Isn't this what C is saying? So, why isn't C the answer?

No, because it is the part of sum which would be the answer.



If you read my whole post, you would understand.

Asked: How many seven-letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (a, e, i, o, u) and the repetition of the letters is allowed?

There will be 4 consonants (21 choices each) and 3 vowels (5 choices each) and arrangement of 7 letters.

Case 1: 4 consonants are same and 3 vowels are same.
Number of seven-letter words = 21^4*5^3*7!/4!3!

Case 2: 4 consonants are all different and 3 vowels are also all different
Number of seven-letter words = 21^4*5^3*7!

IMO E

I think the answer to this question should be C. Here is my thinking

-- Select any 3 places in the seven letter word 7C3
--In these 3 places, each has 5 possible vowels so 5^3
--In the rest 4 places, each has 21 possible consonants so 21^4

Total number of possible words is 7C3*(21^4)*(5^3)

Now the above also includes the cases where all the consonants are different, or only some of them same and others different

Request you to please let me know if there is any flaw in my solution

I think there is a catch people are sturbonly clinging to the possibility that there can be only 3 vowels , even though it's stated repition is allowed and no where is it written that the vowels cannot be more than 3, 'if each word contains 3 vowels' this doesn't imply the vowels count has to be strictly 3

THerefore
5*5*5*26*26*26 = 5^3*26^3
Since i placed 3 vowels on the fore and the remanining all the alphabhets are chosen

Anyway IMO E