Permutation question.

• June 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...

But, according to the book the answer isn't matching...it starts with 2.......000, sorry, don't have the answer list, lost it(Crying), but still. please help..
• June 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...

But, according to the book the answer isn't matching...it starts with 2.......000, sorry, don't have the answer list, lost it(Crying), but still. please help..

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

• June 15th 2009, 02:11 PM
Plato
This is another way to get the answer GrandDad gave.
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: $\left(^{10}\mathcal{P}_4\right)(7!)=25401600$
• June 16th 2009, 03:42 AM
ice_syncer
Quote:

Originally Posted by Plato
This is another way to get the answer GrandDad gave.
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??
• June 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'.
• June 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!