# 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 $24\times 120=2880$.

• Why 4x6 and not 4+6, please explain. I can't wrap my head around it. Thanks. – chanzerre Sep 6 '16 at 12:45
• Explanation added. Hope this helps. – pi66 Sep 6 '16 at 12:48

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

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