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18 Apr 2019, 01:14
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How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?
Bookish Solution:
In the word INVOLUTE, there are 4 vowels, namely, I,O,E,U and 4 consonants, namely, N, V, L and T.
The number of ways of selecting 3 vowels out of 4 = 4C3 = 4.
The number of ways of selecting 2 consonants out of 4 = 4C2 = 6.
Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24.
Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880.
My doubt:
I tried solving this question using the concept of "Permutation" as the order of letters is important.
4P3 X 4P2 (taking 03 vowels out of 04 and 02 vowels out of 04) = 288
I think I have got some concept gap while applying Permutation to solve the question. That's why my answer is incorrect.
Can anyone please point out the gap & solve the question using permutation.
Bookish Solution:
In the word INVOLUTE, there are 4 vowels, namely, I,O,E,U and 4 consonants, namely, N, V, L and T.
The number of ways of selecting 3 vowels out of 4 = 4C3 = 4.
The number of ways of selecting 2 consonants out of 4 = 4C2 = 6.
Therefore, the number of combinations of 3 vowels and 2 consonants is 4 × 6 = 24.
Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is 24 × 5 ! = 2880.
My doubt:
I tried solving this question using the concept of "Permutation" as the order of letters is important.
4P3 X 4P2 (taking 03 vowels out of 04 and 02 vowels out of 04) = 288
I think I have got some concept gap while applying Permutation to solve the question. That's why my answer is incorrect.
Can anyone please point out the gap & solve the question using permutation.
18 Apr 2019, 18:43
Kudos
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There are 4 vowels (I, O, U and E) and 4 consonants (N, V, L, and T) in the word INVOLUTE. So there are 4C3 = 4 ways to choose 3 vowels and 4C2 = 6 ways to choose 2 consonants. However, once we have chosen a particular set of 3 vowels and 2 consonants, there 5! ways to arrange them to form words (with or without meaning). Therefore, there are a total of 4 x 6 x 5! = 2880 words can be formed.
The reason 4P3 x 4P2 = 288 is yielding a wrong answer is because that figure is the number of words where three vowels are followed by two consonants, whereas the question is asking for words involving 3 vowels and 2 consonants in any order.
The reason 4P3 x 4P2 = 288 is yielding a wrong answer is because that figure is the number of words where three vowels are followed by two consonants, whereas the question is asking for words involving 3 vowels and 2 consonants in any order.
