1
$\begingroup$

How many words can be formed, each of $2$ vowels and $3$ consonants from letters of the word "DAUGHTER"

What my textbook has done: it has first taken combinations of vowels and then consonants then multiplied them altogether. Now for each combination of words they can be shuffled in $5!$ ways, so multipliying by $5!$ we get the required answer.

My question is: why has the book used combinations instead of permutations while selecting vowels and consonants?

Thanks.

$\endgroup$
5
$\begingroup$

All the letters are different, so that makes things easier.

Pick the two vowels ($_3C_2$) and pick the three consonants ($_5C_3$) and then pick what order they go in $(5!)$. So the answer is $3 \cdot 10 \cdot 120 = 3600.$

You take combinations of the vowels and consonants because the order of them doesn't matter at that point. You order them in the last step, after you've chosen which ones go in your five-letter word.

In other words, it doesn't matter that I pick $A$, then $U$, instead of $U$, then $A$. It just matters that I picked the set $(A,U)$.

$\endgroup$
  • $\begingroup$ But i am trying do first by taking permutations ,i am trying for shorter words ,but coudn;t figured logic yet $\endgroup$ – Taylor Ted Jun 4 '15 at 19:13
  • $\begingroup$ I suppose you can do it that way, but it becomes quite a bit harder to count things correctly. Does the "combinations-first" way make sense, though? $\endgroup$ – John Jun 4 '15 at 19:20
1
$\begingroup$

Your query why not permutation first ? As, you have to make words of length=$5$. And of these $5$, $2$ are vowels and $3$ consonants. Since, you have to first get those $2$ vowels and $3$ consonants to make the desired word. So first operation has to be combination(selection operation), which will select $2$ vowels out of $3$ vowels(A,E,U) and then you have to select 3 consonants out of $5$(D,G,H,T,R). And they need to be multiplied, as there can be many such combinations i.e $C(3,2)*C(5,3)$. Now that you have formed $5$ letter word. These letters can be arranged among themselves to make different words. Hence, you need to apply permutation(arrangement) i.e. $5!$, making final result= $C(3,2)*C(5,3)*5!$.

$\endgroup$

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.