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.
The word is 'INVOLUTE'
Number of consonants = 4
Number of vowels = 4.
The words formed should contain 3 vowels and 2 consonants.
The problems becomes:
[i] Select 3 vowels out of 4.
[ii] Select two consonants out of 4. Number of selections =
[iii] Arrange the five letters [3 vowels + 2 consonants] to form words.
Number of permutations = 5!
[iv] Apply fundamental principle of counting:
Nội dung chính Show
- Permutations and Combinations
- Mathematics
- How many words with or without meaning each of 3 vowels and 2 consonants can?
- How many words each of 3 vowels and 2 consonants can be formed from the letters of the word equation?
- How many words each of 3 vowels and 2 consonants can be formed from the letters of the word daughter?
- How many words may be formed with 3 consonants and 2 vowels so that no two consonants remain together?
Number of words formed =
=
= 4 x 6 x 120 = 2880
Hence, the number of words formed = 2880
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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.
How many words with or without meaning each of 3 vowels and 2 consonants can?
Required number of ways =2880.
How many words each of 3 vowels and 2 consonants can be formed from the letters of the word equation?
Therefore, the total number of words can be formed is = [4C3 X 4C2 X 5!] = 2880.
How many words each of 3 vowels and 2 consonants can be formed from the letters of the word daughter?
Therefore, 30 words can be formed from the letters of the word DAUGHTER each containing 2 vowels and 3 consonants. Note: A Permutation is arranging the objects in order.
How many words may be formed with 3 consonants and 2 vowels so that no two consonants remain together?
Number of groups, each having 3 consonants and 2 vowels = 210. Each group contains 5 letters. = 5! = 120.