Question

how many words can be formed from LGBASALLOTE? words not necessarily to be in the dictionary.. i dont suppose this is a simple permutation problem? and yeah im narcissistic so i always use my name :p

Answer 1

Letters:
L 3 of 'em
A 2
G 1
B 1
S 1
O 1
T 1
E 1
So, 11 total.

Counting 11 character words ain't so bad.
Let's place the Ls and the As first.
Number of ways to place the Ls:
11 choose 3
Now that I've chosen the Ls, the number of ways to place the As:
8 choose 2
Now that I've chosen the Ls and As, I need only choose a letter for the remaining 6 spots which is
6!
(11C3)*(8C2)*(6!)

Answer 2

the C means...?

Answer 3

"Choose"
Combinations.
\[(11C3) = \frac{11!}{(11-3)!3!}\]

Answer 4

so how do i interpret the expression you had in the \(\frac{n!}{r!}\) form?

Answer 5

You want this expanded?
(11C3)*(8C2)*(6!)

Answer 6

\[\frac{11!}{(11-3)!3!}\frac{8!}{(8-2)!2!}6!\]

Answer 7

wait wait wait...isnt \[\large \frac{11!}{8!3!} = 1??\]

Answer 8

oh wait no sorry

Answer 9

but let's say i choose A first will it be the same as \[\Large \frac{11!}{9!2!} \times \frac{8!}{5!3!} \times 6!\]

Answer 10

the second numerator is supposed to be 9!

Answer 11

Right. And 6!3! in the denominator

Answer 12

right..exactly what i meant haha thanks

Answer 13

Cool, so yeah. That gives us all of the possible 11 character long words...
it doesn't tell us all of the possibilities for shorter words, which becomes a pretty complex problem. Doable, but a bit tedious.

Answer 14

11!/2!3! ?

Answer 15

wait...the possible 11 character words...isnt it supposed to be \[\large \frac{11!}{3!2!}\]?

Answer 16

oh lol it is...

Answer 17

i thought that was the possible number of words...that was just an expanded form of the number of how many times my name can be arranged :p

Answer 18

Hmm okay. Yeah, that's a somewhat more elegant way to calculate it.

Answer 19

"LGBASALLOTE" has 11 letters among which 3 L's and 2 A's the rest are all different.

Total number of permutation possible taken all at a time: \[ \frac{11!}{3!\times 2!}\]

Answer 20

well i knew about the part of the 11 characters...no way to tell for the smaller characters?

Answer 21

Only 10 Characters:
Option 1: only 2 Ls-
\[\frac{10!}{2!2!}\]
Option 2: only 1 A-
\[\frac{10!}{3!}\]
Option 3: 3Ls 2As, 1 of each other letter besides 1
\[\frac{10!}{3!2!}*6\]

Add those up to get the number of 10 char words.

Answer 22

ugh o.O that's enough to show me it wont be simple :p lol

Answer 23

Do you follow the logic there?

Answer 24

somehow...

Answer 25

Haha okay good.