How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE? [duplicate]

Question

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.

Answer 1

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 $24\times 120=2880$.

Answer 2

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