1. ## combinations

The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
How many possible selections are there of four letters.

I did it the long way by coming up with each case. For example,
case 1 - no 's'
P.O.E.E
case 2 - one 's'
S.P.O.E; S.P.E.E; S.O.E.E
case 3 - two 's'
S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
case 4 - three 's'
S.S.S.E; S.S.S.P; S.S.S.O
case 5 - four 's'
S.S.S.S
This adds up to be 12.
Is there a better way to do this??

Thanks.

2. ## Re :

Originally Posted by woollybull
The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
How many possible selections are there of four letters.

I did it the long way by coming up with each case. For example,
case 1 - no 's'
P.O.E.E
case 2 - one 's'
S.P.O.E; S.P.E.E; S.O.E.E
case 3 - two 's'
S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
case 4 - three 's'
S.S.S.E; S.S.S.P; S.S.S.O
case 5 - four 's'
S.S.S.S
This adds up to be 12.
Is there a better way to do this??

Thanks.
i misintepreted , plato has corrected ..

3. Originally Posted by woollybull
The letters P.O.S.S.E.S.S.E.S are shuffled and four selected and placed in a line.
How many possible selections are there of four letters.

I did it the long way by coming up with each case. For example,
case 1 - no 's'
P.O.E.E
case 2 - one 's'
S.P.O.E; S.P.E.E; S.O.E.E
case 3 - two 's'
S.S.P.E; S.S.E.E; S.S.O.E; S.S.P.O
case 4 - three 's'
S.S.S.E; S.S.S.P; S.S.S.O
case 5 - four 's'
S.S.S.S
This adds up to be 12.
Is there a better way to do this??
It may not be better but it is quicker: $\frac{4!}{2!} = 12$

That answer is just for no s’s.Other cases are: Exactly one s: <SPOE>, <SPEE>, <SOEE>
Exactly two s’s: <SSOE>, <SSEE>, <SSOP>, <SSEP>
Exactly three s’: ?
Exactly four s’: ?

4. I don't have easy internet access so I'm trying to interpret what I missed.
Is there a better way?
Also I am after combinations - not permutations (arrangements).
Thanks.

5. Hello, woollybull!

The letters of $POSSESSES$ are shuffled.
Four are selected and placed in a line.
How many possible selections are there of four letters?
Since they are "placed in a line", I assume that their order is important.
. . That is, we are spelling four-letter "words."

I see no way avoid a brute-force listing . . .

$\begin{array}{ccccc}
\text{No S's} \\ \hline \\[-4mm]
\text{Letters: }P, O, E, E & &\frac{4!}{2!} &=& 12\text{ words}\end{array}$

$\begin{array}{ccccc}
\text{One S} \\ \hline \\[-4mm]
\text{Letters: }P, O, E, S & & 4! &=& 24\text{ words} \\
\text{Letters: }P, E, E, S & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]
\text{Letters: }O, E, E, S & & \frac{4!}{2!}&=& 12\text{ words} \end{array}$

$\begin{array}{ccccc}
\text{Two S's} \\ \hline \\[-4mm]
\text{Letters: P, O, S, S} & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]
\text{Letters: }P, E, S, S & & \frac{4!}{2!}&=& 12\text{ words} \\ \\[-4mm]
\text{Letters: }O, E, S, S & & \frac{4!}{2!} &=& 12\text{ words} \\ \\[-4mm]
\text{Letters: }E, E, S, S & & \frac{4!}{2!\,2!} &=& 6\text{ words} \end{array}$

$\begin{array}{ccccc}
\text{Three S's} \\ \hline \\[-4mm]
\text{Letters: }P, S, S, S & & \frac{4!}{3!} &=& 4\text{ words} \\ \\[-4mm]
\text{Letters: }O, S, S, S & & \frac{4!}{3!} &=& 4\text{ words} \\ \\[-4mm]
\text{Letters: }E, S. S. S & & \frac{4!}{3!} &=& 4\text{ words} \end{array}$

$\begin{array}{ccccc}
\text{Four S's} \\ \hline \\[-4mm]
\text{Letters: }S, S, S, S & & \frac{4!}{4!} & =& 1\text{ word} \end{array}$

Now add them up . . .