In response to Permutations and Combinations Thread Question (1 Viewer)

[bored of sc]

Um, the thread wasn't working so I created this one.

Um, you should mulitply your answer by four cause there are 4 possible places where you can have two vowels in succession.

Hope that's right!

 

[bored of sc]

P.S I haven't done this topic in depth yet - so I am probably wrong!

 

[Finx]

Ahh thanks =]

I'd deleted the post because I thought this was the wrong place to post it.

Though the 'Maths Extension 1' board isn't very responsive =[

Thanks again!

 

[bored of sc]

I am guessing you worked out your error in working and took the thread off!

 

[Sarah182]

I have a permutations and combinations question,
In my test on friday the question was "How many ways can the word ALGEBRAIC be arranged" Now the only thing that stumped me was there is two A's. I think I did it right but I'm not sure. Can someone give me hand please?

 

[Finx]

Hmm. I remembered how to do this before, but now I've forgotten. Damnit >_<

 

[bored of sc]

you count both A's as one kinda thing - so there can't be A1 and A2 - well that's what I think anyway

 

[Leo 100]

yeah
there are two possibilities of an A occuring but the sample space is (ALGEBRAIC) not 9 but 8 cause you dont count A twice



.... or something like that, i think... know what i mean?

 

[Finx]

What would the equation look like?

 

[Sarah182]

Well I think I did 8 factorial (8!) because I counted A as one thing and I wasnt sure whether to multiply that by 2! because there are 2 possibilities of A occuring if that makes sense.
In the end I panicked and just did 8! but I'm pretty sure thats wrong.

 

[Finx]

Could I have an explanation on how to do these kinds of permutations please? I don't think I'm very confident with these types of questions.

Using Sarah's guess (8!*2!), I looked up a question asking how many arrangements are in 'ALAN'. Since there are 3 different letters (A, L and N) and two of the same letter, I typed 3!*2! and got the correct answer.

However, for something like 'SNEEZE', there are 4 different letters (S, N, E and Z) and three of the same letters. After typing 4!*3!, the answer was not correct.

Help please =[

 

[Finx]

I figured it out.

You need to find the total possibilities of arrangement using the total number of letters (regardless of whether they are the same as each other or not) then divide by how many of the same letter there are.

So for ALGEBRAIC, you have 9 total letters with 2 of the same letter. Therefore the equation is 9! / 2!

 

[Sarah182]

I figured it out.

You need to find the total possibilities of arrangement using the total number of letters (regardless of whether they are the same as each other or not) then divide by how many of the same letter there are.

So for ALGEBRAIC, you have 9 total letters with 2 of the same letter. Therefore the equation is 9! / 2!

Thanks, I'll know how to do it in my half yearlys now

 

[bored of sc]

don't forget the product rule - just multiplying each event

P.S that's the only thing I have done for that topic in extension maths so far

 

[bored of sc]

yeah
there are two possibilities of an A occuring but the sample space is (ALGEBRAIC) not 9 but 8 cause you dont count A twice



.... or something like that, i think... know what i mean?

yer, i think - all the stuff I am saying is merely logic and reasoning cause I haven't done this factorial stuff yet