Question

How many possible 4 letter words can be made from the word "LGBASALLOTE"?

Answer 1

real words?

Answer 2

just possible 4 letter arrangements

Answer 3

and it's actually much harder than it seems...

Answer 4

\[\frac{ 11! }{ 3! * 2! }\]

Answer 5

that's the permutation for the whole word...

Answer 6

i would assume cases are needed ehre

Answer 7

First you use \[
\left( \begin{array}{c}
11 \\
4
\end{array} \right)
\]To get the total number of words that you could get if there were no indistinguishable letters. Then you have to figure out a way to get rid of double counting.

Answer 8

hmm...i get that first part....

Answer 9

http://openstudy.com/study#/updates/5049bf6fe4b08356fb2c4e07

Answer 10

similar question

Answer 11

that's an easy one...it just has 2 repeating characters.

Answer 12

but what if there are more?

Answer 13

take different cases..

Answer 14

For every word that uses two A, it won't be double counted, but if it uses one A, it will be double counted.

Answer 15

If it has 1 L, then it will be tripled counted, if it has two Ls, it will be counted 3 chose 2 times, if it has 3 L it won't be double counted

Answer 16

There isn't necessarily a nice looking formula for this.

Answer 17

\[
\left( \begin{array}{c}
11 \\
4
\end{array} \right)
-
\left( \begin{array}{c}
3 \\
1
\end{array} \right)

\left( \begin{array}{c}
8-3 \\
3
\end{array} \right)
-
\left( \begin{array}{c}
3 \\
2
\end{array} \right)

\left( \begin{array}{c}
8-3 \\
2
\end{array} \right)
-
\left( \begin{array}{c}
3 \\
3
\end{array} \right)

\left( \begin{array}{c}
8-3 \\
1
\end{array} \right)
\]\[-
\left( \begin{array}{c}
2 \\
1
\end{array} \right)

\left( \begin{array}{c}
8-2 \\
3
\end{array} \right)
-
\left( \begin{array}{c}
2 \\
2
\end{array} \right)

\left( \begin{array}{c}
8-2 \\
2
\end{array} \right)
\]Something like this...

Answer 18

Though I might be overlooking something.

Answer 19

....wow.....

Answer 20

Those 8s should be 11s

Answer 21

why subtraction?

Answer 22

It's my best attempt at trying to find any overlap.

Answer 23

LGBASALLOTE
There are 3 L's and 2 A's, nothing else is repeated. Getting rid of the repeated letters for now...
LGBASOTE
For 4 letter words with no repeated letters, there are
8P4 possibilities.
Suppose there are 2 L's and no other repetitions
There are 6 (4C2) ways to scatter 2 L's around a 4 letter word
times the number of ways the 7 non-L letters could be scattered around the two remaining "slots" So that's
6 x 7P2
Suppose there are 2 A's and no other repetitions
There are 6 (4C2) ways to scatter 2 A's around a 4 letter word
times the number of ways the 7 non-A letters could be scattered around the two remaining "slots" so that's
6 x 7P2
Suppose there are 3 L's (of course, no other repetition may occur, as there is only one more letter)
There are 4 (4C3) ways to scatter 3 L's around a 4 letter word
times the number of ways the 7 non-L letters could be placed in the remaining "slot" so that's
4 x 7P1 = 4x7 = 28.
Suppose there are 2 L's and 2 A's
Then the number of ways they can be scattered is equal to the number of ways two L's can be scattered around a 4 letter word
so that's
6.
TOTAL
8P4 + (6)(7P2) + (6)(7P2) + 28 + 6

Answer 24

why 8P4?

Answer 25

shouldn't it be 6P4?

Answer 26

Because there are 8 distinct letters?

Answer 27

so you didn't consider the arrangement where there are no L or A?

Answer 28

No I didn't... Thanks for the heads-up
Recalculating :D

Answer 29

same thing for @wio why 11C4?

Answer 30

Wait, 8P4 already does account for the possibilities that there may be no L or A.

Answer 31

it does?

Answer 32

I think so. You simply cannot use all 8 letters in a 4-letter word, after all.

Answer 33

hmm...i don't seem to find any mistake in that solution....

Answer 34

except for that last 6....

Answer 35

shouldn't it be \[\huge _3 C _ 2 \times _2C_2 \times \frac{4!}{2!2!}\]

Answer 36

making it equal to 18....

Answer 37

The last 6? When the 4 letter word consists purely of 2 L's and 2 A's?

Answer 38

yes

Answer 39

Well, it's only 6, after all, brute force gives us
LLAA
LALA
LAAL
ALLA
ALAL
AALL

Answer 40

yes...that's the possible way of arranging...but what about the possible ways of picking the L?

Answer 41

doesn't that make a difference?

Answer 42

It shouldn't, unless you mean to say, for instance
LGBA can be counted three times depending on which "L" I used...
If it does, then it basically means all my calculations were near-useless...

Answer 43

.....i suppose i can't prove you wrong then....

Answer 44

^Which doesn't necessarily mean I'm right LOL

Answer 45

8C4 *4! + 8C3*4!/2! + 8C2 * 4!/3! + 8C3*4!/2! +8C2 *4!/2!2!