# Permutation question.

• Jun 15th 2009, 11:26 AM
ice_syncer
Permutation question.
How many arrangements can we make from the letters from the word PERMUTATIONS such that there are always 4 letters in between P and S?

I think it is ((superscript)10P(subscript)4)/2! * 6! because since P and S are utilised there are only 10 letters left and , since only 4 letters should be there in between P and S, there are 4 slots, so, the number of possible arrangements between P and S is 10*9*8*7 but, T repeats twice...so it hsuld be 10P4/2! and, this "object" between P and S can be arranged in 6 ways within the huge word...

• Jun 15th 2009, 11:59 AM
Permutations
Hello ice_syncer
Quote:

Originally Posted by ice_syncer
How many arrangements can we make from the letters from the word PERMUTATIONS such that there are always 4 letters in between P and S?

I think it is ((superscript)10P(subscript)4)/2! * 6! because since P and S are utilised there are only 10 letters left and , since only 4 letters should be there in between P and S, there are 4 slots, so, the number of possible arrangements between P and S is 10*9*8*7 but, T repeats twice...so it hsuld be 10P4/2! and, this "object" between P and S can be arranged in 6 ways within the huge word...

Ignore for now the repeated T's, and write the P down first (anywhere).

Then the S can be written down in 2 ways - to the left or to the right of the P.

The four spaces between them can be filled in $10\times 9\times 8 \times 7$ ways.

Then this block of six letters can be arranged with the remaining 6 individual letters in 7! ways (not 6!).

The total now is $2 \times 10\times 9\times 8 \times 7 \times 7!$. But we now have to divide by 2! because of the repetition of the T's.

Answer: $10\times 9\times 8 \times 7 \times 7!=25401600$

• Jun 15th 2009, 02:11 PM
Plato
With subscripts $PERMUT_1 AT_2 ION$ we have twelve different letters.
We have $^{10}\mathcal{P}_4$ possible strings to go between P & S.
We double that to care for P____S or S____P.
Now we have seven blocks to arrange: $7!$.
Now divide by two to remove the subscripts.

Answer: $\left(^{10}\mathcal{P}_4\right)(7!)=25401600$
• Jun 16th 2009, 03:42 AM
ice_syncer
Quote:

Originally Posted by Plato
With subscripts $PERMUT_1 AT_2 ION$ we have twelve different letters.
We have $^{10}\mathcal{P}_4$ possible strings to go between P & S.
We double that to care for P____S or S____P.
Now we have seven blocks to arrange: $7!$.
Now divide by two to remove the subscripts.

Answer: $\left(^{10}\mathcal{P}_4\right)(7!)=25401600$

OK, but I don't understand how it is 7! instead of 6!..how come 7 slots? isnt' it 6 slots??
• Jun 16th 2009, 04:39 AM
Plato
Quote:

Originally Posted by ice_syncer
OK, but I don't understand how it is 7! instead of 6!..how come 7 slots? isnt' it 6 slots??

Example:
$\underbrace {\boxed{PRTUAS}}_1$ $\underbrace {\boxed{N}}_2$ $\underbrace {\boxed{M}}_3$ $\underbrace {\boxed{T}}_4$ $\underbrace {\boxed{O}}_5$ $\underbrace {\boxed{I}}_6$ $\underbrace {\boxed{E}}_7$

As you can clearly see there seven 'blocks' or 'slots'.
• Jun 16th 2009, 05:09 AM
ice_syncer
Quote:

Originally Posted by Plato
Example:
$\underbrace {\boxed{PRTUAS}}_1$ $\underbrace {\boxed{N}}_2$ $\underbrace {\boxed{M}}_3$ $\underbrace {\boxed{T}}_4$ $\underbrace {\boxed{O}}_5$ $\underbrace {\boxed{I}}_6$ $\underbrace {\boxed{E}}_7$

As you can clearly see there seven 'blocks' or 'slots'.

Oh I see!, thanks!