Question

how many three letter words can be formed from LGBASALLOTE? note: words not necessary to be in the dictionary

Answer 1

is it \[\large \frac{11!}{8!}\]

Answer 2

11x10x9 , maybe

Answer 3

You'd divide off each individual factorial for the number of repeated letters
Like there are 3 L's so divide off 3!
There are 2 A's so divide off 2! as well

etc

although i guess it depends on how you define a word, just as a group of letters or has to have a vowel or something. D:

Answer 4

it is 11!/8!. I just did a very similar problem. The equation to use is (n!)/(n-r)!

Answer 5

but that's the number of how many times LGBASALLOTE can be arranged @AccessDenied

Answer 6

thanks :D

Answer 7

Hmm... where does the 8! come from?

Answer 8

11 - 3

Answer 9

oh wait formula ic

Answer 10

Hmm... 11! / 8! = 11x10x9
Since I'm not good at using ''!''
I usually just think it in a VERY traditional way...
there are 11 letters to choose for the 1st letter of the word, 10 letters to choose for the 2nd letter of the word, 9 letters to choose for the 3rd letter of the word
So, multiply them all... And get the answer
*I wish someone can teach me these stuffs :( *

Answer 11

OH okay. I see where this is coming from. D:

My mind just read problem and said "how many ways can you arrange it uniquely" automatically lol.

Answer 12

i hear you girl..im also trying to self-learn these stuffs @_@

Answer 13

Hmm.. but then would you have to consider non-unique cases, like choosing ASA and then the other A first for ASA?

Answer 14

that...is a good point

Answer 15

hmm? That has been taken into account already?!
Since I haven't specified which one to choose ...

Answer 16

maybe it should be \[\large \frac{11!}{8!3!2!}\]

Answer 17

like, in your case you consider all letters as unique; like, you choose A or the other A it's considered as different cases

But if you consider ASA and ASA as the same word, you'd have to eliminate the repeats. D:

Answer 18

This gives a lot of really useful information - http://www.mathsisfun.com/combinatorics/combinations-permutations.html

Answer 19

My English fails me.... :S

Answer 20

so \[\frac{6!}{8!}\] wouldnt make sense >.<

Answer 21

None of the answer posted is correct :)

Answer 22

No That does't make sense @lgbasallote

Answer 23

Wow~~ FFM comes to save us!!!

Answer 24

11! / 8! 3! 2!

11*9*8
------
6*2

11*3*2

?

Answer 25

Nopes, The word "L G B A S A L L O T E" has 3 L's and 2 A's the rest are all different.

Answer 26

\[\large \frac{6!}{3!}\] seems few...

Answer 27

There is no easy way to solve this one. I will do it with generating functions.

Answer 28

oh..so this isnt simple :/ these things are hard to see if they are difficult or not..

Answer 29

wait i didn't even do my math correctly. blah. ;[
well, i guess its no use if it's not that anyways. lol

Answer 30

If you guys want I can post the generating function approach.

Answer 31

I'm personally curious about it myself. :)

Answer 32

Go~~~

Answer 33

The coefficient of \(x^3\) in:
\[3! \times (1+x)^6 \times \left(1+x+\frac {x^2}{2!} \right )\times \left(1+x+\frac {x^2}{2!} +\frac{x^3}{3!}\right )\]


If my algebra is correct the answer is \(379\)

Answer 34

I get 379 as well when I worked it out on paper. Very strange method to me, so I guess I have to study more into the subject later to understand better. D:

Thanks, FoolForMath! :D

Answer 35

Its' is actually very easy, if you are interested I will show you how to understand and derive it on your own.

Answer 36

Hmm, that'd be really interesting. :D
Like, I can kind of see some patterns in the expression but not completely obvious how it all works out.

Answer 37

First read about Generating function from here :
http://users.softlab.ntua.gr/~nidal/discrete/notes/examples/gen2.pdf

and here:

http://courses.csail.mit.edu/6.042/fall05/ln11.pdf

and then read this answer:

http://math.stackexchange.com/questions/20238/

Answer 38

Take it slow, and enjoy the ride :)