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    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, but still. please help..
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    Permutations

    Hello ice_syncer
    Quote Originally Posted by ice_syncer View Post
    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, 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

    Grandad
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    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
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    Quote Originally Posted by Plato View Post
    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??
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    Quote Originally Posted by ice_syncer View Post
    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'.
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  6. #6
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    Quote Originally Posted by Plato View Post
    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!
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