How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word ‘INVOLUTE’?
How many words, each of 3 vowels and 2 consonants, can be formed from the letters of the word ‘INVOLUTE’?
Answer : In the word ‘INVOLUTE’ there are 4 vowels, ‘I’,’O’,’U’ and ‘E’ and there are 4 consonants, ‘N’,’V’,’L’ and ‘T’. 3 vowels out of 4 vowels can be chosen in 4C3ways. 2 consonants out of 4 consonants can be chosen in 4C2 ways. Length of the formed words will be [3 + 2] = 5. So, the 5 letters can be written in 5! Ways. Therefore, the total number of words can be formed is = [4C3 X 4C2 X 5!] = 2880.
My 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 $4 = C[4,3] = 4$. The number of ways of selecting $2$ consonants out of $4 = C[4,2] = 6$. Therefore, the number of combinations of $3$ vowels and $2$ consonants is $4+6=10$.
Now, each of these $10$ combinations has $5$ letters which can be arranged among themselves in $5!$ ways. Therefore, the required number of different words is $10\times5! = 1200$.
But the answer is $2880$.
What am I doing wrong? Please explain.
Solution
I N V O L U T E
Number of letters = 8
Vowels = I, O, U, E
Consonants = N, V, L, T,
Number of ways to select 3 vowels = 4C3
Number of ways to select 2 consonants = 4C2
Number of ways to arrange these five letters
= 4C3×4C2×5!
=4×6×5×4×3×2×1
=2880
Required number of ways =2880
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Joined: 27 Nov 2018
Posts: 63
How many words, with or without meaning, each of 3 vowels and 2 conson [#permalink]
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.
Target Test Prep Representative
Joined: 14 Oct 2015
Status:Founder & CEO
Affiliations: Target Test Prep
Posts: 16181
Location: United States [CA]
Re: How many words, with or without meaning, each of 3 vowels and 2 conson
[#permalink]
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.
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Re: How many words, with or without meaning, each of 3 vowels and 2 conson [#permalink]
18 Apr 2019, 17:43
Moderator:
Retired Moderator
994 posts
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= 120 ways.