Hello everyone! It's me again, Francis, and I'd like to ask for your help regarding another math problem, this time from combinatorics. I have a solution in mind, and though it's not as elegant as I want it to be, I hope you see my reasoning. Feel free to comment; anything that helps is welcome. Thank you very much!
In how many ways can the letters of the word “Manileño” be arranged such that the vowels are arranged in alphabetical order from left to right?
First, write the vowels in the proper order, with spaces before, after, and in between for the consonants.
A E I O
Now let us place the consonants on the spaces in between. Note that each space may contain any number of consonants, ranging from none to a maximum of four.
We have 4 choices for the first consonant and 5 choices for its position. Also, we have 3 choices for the second consonant (because the first has already been used), but there are still 5 positions available. Similarly, there are 2 and 1 choices for the third and fourth consonants, respectively, with 5 available positions for each. Multiplying these numbers results in
4 ∙ 5 ∙ 3 ∙ 5 ∙ 2 ∙ 5 ∙ 1 ∙ 5=5^4 ∙ (4!)=625 ∙ 24=15 000
possible arrangements of the eight letters.
Objection: But this solution fails to consider the fact that the consonants may be arranged in any order between the vowels.
Rebuttal: Yes it does. This is why we used 4! as one of the factors of our answer. For example, if L and Ñ are the first and second chosen consonants, respectively, and if they are both placed between A and E, they will only appear as ALÑE and not as AÑLE because the order of the consonants was already determined before placing them between the vowels.
PS Sorry, no LaTeX again. I just pasted the whole thing from MS Word because I sent my solution to my friends. Unfortunately they all had different answers. Oh well. Happy solving!