1
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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 $24\times 120=2880$.

$\endgroup$
  • $\begingroup$ Why 4x6 and not 4+6, please explain. I can't wrap my head around it. Thanks. $\endgroup$ – chanzerre Sep 6 '16 at 12:45
  • $\begingroup$ Explanation added. Hope this helps. $\endgroup$ – pi66 Sep 6 '16 at 12:48
-1
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. The egregious capitalization is going to invite down votes. For formulation, please use MathJax. $\endgroup$ – dantopa Feb 1 at 16:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.