How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE? 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 number of combinations of $3$ vowels and $2$ consonants should be $4\times 6=24$ instead of $4+6=10$. Since you are considering combinations, each set of $3$ vowels and each set of $2$ consonants form a combination, so you need to multiply and not add.

The combinations are $(\{I,O,E\},\{N,V\}),(\{I,O,E\},\{N,L\}),\dots$. You can try writing out all $24$ combinations as an exercise.

Then you will get $24\times 120=2880$.
IN THE WORD INVOLUTE, WE HAVE 4 VOWELS AND 4 CONSONANTS.

OUT OF 4 VOWELS WE CHOOSE 3 VOWELS THAT MAKES IT 4C3.
OUT OF 4 CONSONANTS WE CHOOSE 2 CONSONANTS THAT MAKES IT 4C2.

Therefore on calculation, we get

4C2*4C3*5!( we take 5! because we can arrange it in 5 factorial ways.)

Therefore,
6*4*5!= 24*120=2880