AJ0784 wrote:
Bunuel wrote:
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?
A. \(21C4*5C3*7!\)
B. \(21^4*5^3\)
C. \(21^4*5^3*\frac{7!}{4!*3!}\)
D. \(21^4*5^3*7!\)
E. None of these
How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)
a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these
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.