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.
[iii] Arrange the five letters [3 vowels + 2 consonants] to form words.
Number of permutations = 5!
[iv] Apply fundamental principle of counting:
Number of words formed =
=
= 4 x 6 x 120 = 2880
Hence, the number of words formed = 2880
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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.