# How many words can be formed using all the letters of “DAUGHTER” so that vowels always come together?

How many words can be formed using all the letters of "DAUGHTER" so that vowels always come together?

I understood that there are 6 letters if we consider "AUE" as a single letter and answer would be 6!. Again for AUE it is 3!, but I didn't get why to do 6! * 3!.

Can't we just add (6! + 3!) to get final result?

• You multiply because you are doing a permutation of your 6 "letters" AND the permutation of the 3 letters in your "AUE". Generally, if it was an OR, and not an AND, you would add them. This revolves around the rule of product (en.wikipedia.org/wiki/Rule_of_product) where you are doing two things at the same time, not separately. – anakhro Jul 22 '15 at 12:52
• Here is an answer referring to the notion of cartesian product: math.stackexchange.com/questions/1362768/… . Maybe it is of help. – Christian Blatter Jul 22 '15 at 13:50

Imagine your three vowels as a block. Ignore the ordering of the block for the moment. Now you have 6 remaining "letters"- the original consonants plus this "vowel block". Now, there are $6$ of these altogether. So you get $6!$ ways of ordering them- which is the standard permutation formula. Now for $each$ of these orderings, you can internally order the three vowels in the block. So you multiply by $3!$, which is the number of ways to order the vowels.

EDIT: The intuition for multiplication can be strengthened with some visualisation. Take a sheet of paper. Imagine all the "external" orderings of consonants and vowel block as being listed vertically. Now for each of these, write horizontally the 6 corresponding "internal" orderings of the vowels. You get a rectangular grid, each cell containing exactly one ordering. The area of this grid is clearly the breadth by its height. Hence $6!3!$.

I see often that people have troubles with the rule of product in combinatorics. I also see often that people who claim to not have troubles with it and try to explain it to the aforementioned clueless people just end up saying "it's just a formula, don't worry about where it came from and why we do it, just memorize it!"--not a good tactic at all.

I will say I myself don't have a deep understanding of it and will at the end of this answer tell you it is just a formula, but I hope I can help you understand the formula a bit more.

Recall from (likely) grade school where you made tree diagrams to express combinatorics problems visually. Take the following simplified question:

How many 'words' can you arrange out of the letters in the set $\{x,y,z\}$?

Our tree diagram will look like this: Now if we count each step in our tree, we see in the first step of our tree (far left), we select one of three letters. Each is a root for its own subtree. That's $3$ trees started. So we have: $$(3\ trees).$$ Now in the second step (middle), we select one of the letters we have left that isn't the original we started with, leaving us with $2$ new choices per tree. So that is: $$(3\ trees) \times (2\ choices) = 6 \ total.$$ We now consider the final step where we choose the remaining letter. However, this step doesn't further split our tree's branches per se, it just extends them. So we get: $$(3\ trees) \times (2\ choices) \times (1\ more\ choice) = 6 \ total,$$ which is identical to our exact formula we had previously memorized, that is: $$3! = 3 \times 2 \times 1 = 6.$$

Now suppose we change around the question a little. Suppose this new condition:

After choosing $z$, choose $z$ to be an element from the set $\{a,b,c\}$.

(imagine this choice is like choosing a different configuration of AUE, but in this case instead of $3!$ choices for the configuration, we have only $3$).

Our tree diagram looks like this: which is quite similar to before (notice we just appended the new choice onto the end since even if we put it directly after every $z$, it would have the same number of branches (try this yourself if you want!)) but with an additional step.

As with the other steps, we just multiplied the number of choices we already had by the number of new choices like so: $$(3\ trees) \times (2\ choices) \times (1\ more\ choice) \times (3\ choices\ for\ z)= 18\ total.$$

So perhaps now you have a better understanding of why we use the rule of product in the case of your question. As you can tell, it still is sort of a formula you have to know. When you have $\alpha$ ways of doing one task and $\beta$ ways of doing another, you then have $\alpha \times \beta$ ways of doing both.

For the case for addition, note very well that it is more associated with the word "or" than with the word "and". So if you want to calculate the number of ways to do $\alpha$ or $\beta$, then you will have to think about using addition, but even then it isn't so straightforward sometimes (google for inclusion-exclusion principle, among other things for more information on it).

• It's a pity this isn't an answer to a question on the meaning of multiplication. I feel it's lost in here. – Colm Bhandal Jul 25 '15 at 11:23
• @ColmBhandal I do not understand what you are saying. – anakhro Jul 25 '15 at 13:17
• It is a very good answer: it deserves to be in a separate thread. – Colm Bhandal Jul 25 '15 at 15:42

Consider two boxes to be filled with coins and we are given 7 different coins. For the first place you can select any of those 7 coins. Let we have chosen first coin for 1st box now you can fill the second box in 6 different ways. So with 1st coin in first box we have 6 permutations. Instead of having filled the first box with fist coin we could have filled it with 2nd coin, again the second box can be filled in 6 different ways. Then 6 ways if we put 3rd coin in first box, then 6 ways if we put 4th coin in first box, then 6 ways if we put 5th coin in first box, the 6 ways if we put 6th coin in first box and finally 6 ways if we put 7th coin in first box. So you see we have (6+6+.. upto 7 times permutations). In general if we can fill first box in $m$ ways and second box in $n$ ways then the total number of ways in which we can make different permutations is $m*n$, this is because for every 1 way out of m ways in which first box can be filled there are $n$ ways to fill the second box. In other words for 1st object in box one there are $n$ possible ways to fill the second box, for 2nd object in first box there are again $n$ possible ways to fill the second box. At last for the mth object in the first box there are $n$ possible ways to fll the second box. This fact is called "Fundamental Theorem of Multiplication".

If we had three boxes where first box could be filled in m, second in n and third in p ways then total no of permutations would be $m\times n \times p$, why? Because From every one possible permutation out of $m*n$ of first two boxes the third can be filled in $p$ different ways.

We can extend this theorem to any number of boxes. Now in place of boxes we can conceive of jobs where 1st box is 1st job which can be done in $m$ ways, 2nd box is 2nd job which can be done in $n$ ways and so on.

You have DAUGHTER splitted to "DGHTR" and "AUE", lets represent a vowel by 0 and another letter by 1

All possible combinations are written as: 11111000 or 11110001 or 11100011 or 11000111 or 10001111 or 00011111

Permutations of vowels counted to 3!

Permutations of letters counted to 5!

Each permutation of letters hold 3! vowel permutations, see cleary that i said each, so there is no union of two probablities here, which the case of a multiplication.

In other hand, a union of two permtations is writen $Perm(x)+Perm(y)-Perm(x)\bigcap perm(y)$

A pemutation of $11..(n)00..(k)11..(l)$ is totally independant from permutations of $111...(n')0..(k')111...(l')$, so $Perm(x)\bigcap perm(y)=0$

$3!5!+3!5!+3!5!+3!5!+3!5!+3!5!=3!*6*5!=3!6!$