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How many arrangements of the word MARMALADE can be made with the vowels in the original order? What is the procedure for doing this problem? Is there more than one way of approaching it?

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3 Answers

up vote 5 down vote accepted

We first change the problem a bit, and deal with the word marMalade.

List the vowels in their original order: a a a e. They determine $5$ "gaps" (we are counting the space before the first a and the space after the $e$ as gaps).

Now look at the letters m r M l d. The first of these, namely m, can be slipped into any of the $5$ gaps. After we do that we have a $5$-letter words, i.e. $6$ "gaps." The letter $r$ can be slipped into any of these $6$ gaps.

Now we have a $6$-letter word, that is, $7$ gaps. We can slip $M$ into any of these. Continue. We end up with $5\cdot 6\cdot 7\cdot 8\cdot 9$ "words."

Now turn the M back into an m. This divides the number of words by $2$. Thus the total number of words for the original problem is $$\frac{5\cdot 6\cdot 7\cdot 8\cdot 9}{2}.$$

There are many other ways to solve the problem.

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Nice way to think about it –  sonicboom Dec 3 '12 at 16:15
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There will be more than one approach. Here is one set of hints:

  • Look at vowels: AAAE. There are four.

  • Look at consonants: MRMLD. There are five including a pair identical.

  • How many ways of arranging the consonants taking account of the fact there is a pair?

  • How many ways of interleaving four vowels and five consonants already in a given order?

Multiply those two together.

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Vowels can be arranged in $v = \frac{4!}{3!1!}$ ways. Consonants can be arranged in $c = \frac{5!}{2!1!1!1!}$ ways. Then multiply $v$ by $c$ and divide by four, we are only in interested one particular arrangement of the vowels. Is that correct? –  csss Dec 3 '12 at 7:58
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Actually, what about this. There are $\frac{9!}{3!2!}$ ways of permuting the letters of MARMALADE. We should divide this be number by $4$ are we are only interested in one permutation of 4 possible permutation of the vowels? –  csss Dec 3 '12 at 8:13
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@csss: Your first comment is wrong, because you don't take into account the ways of interleaving the vowels and consonants (i.e. that way of counting doesn't distinguish between MARMALADE and AMRMALADE). Your second comment is correct (great approach!). You can count it more simply by not dividing by the 3!, and simply writing it as $\frac{9!}{4!2!}$ where the denominator is meant to compensate for overcounting permutations of the two M's, and the 4 vowels. –  ShreevatsaR Dec 3 '12 at 8:35
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An approach would be to consider the vowels as all one letter, such as in $$ \text{MARMALADA} $$ where it is understood that we want to replace $\text{AAAA}\to\text{AAAE}$.

Thus, for each permutation we wish to count there are $4!$ permutations of the $\text{A}$s and $2!$ permutations of the $\text{M}$s that don't get counted to get the $9!$ total permutations of $9$ letters.

Thus, there are $\frac{9!}{4!2!}$ permutations of the letters in $\text{MARMALADA}$ which is the number of permutations of the letters in $\text{MARMALADE}$ that keep the vowels in order.

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