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Thread: permutation when p things are always present

  1. #1
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    permutation when p things are always present

    hi,
    i have a problem in understanding a combinatorics formula in my algebra book.
    hope i am posting in the right section, i found no section on combinatorics,sorry.
    it says that the no of permutations of n different things taken r at a time ,when p particular things are to be always included in each arrengement is
    p! (r-p+1) ((n-p) permutation (r-p))
    i proceeded in the following way-let me first find the arrengements of the things other than the p ones. that is n-p P r-p.
    now the p things can arrenge in p! ways.
    but i cannot understand how (r-p+1) is coming here.
    thanks in advance for any help.
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  2. #2
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    Hello mathlove

    Welcome to Math Help Forum!
    Quote Originally Posted by mathlove View Post
    hi,
    i have a problem in understanding a combinatorics formula in my algebra book.
    hope i am posting in the right section, i found no section on combinatorics,sorry.
    it says that the no of permutations of n different things taken r at a time ,when p particular things are to be always included in each arrengement is
    p! (r-p+1) ((n-p) permutation (r-p))
    i proceeded in the following way-let me first find the arrengements of the things other than the p ones. that is n-p P r-p.
    now the p things can arrenge in p! ways.
    but i cannot understand how (r-p+1) is coming here.
    thanks in advance for any help.
    Are you sure you have copied this formula correctly? I don't see where it comes from either. But neither do I agree with the answer you have suggested!

    Let us imagine that we have $\displaystyle \displaystyle r$ empty boxes to fill, and $\displaystyle \displaystyle p$ of them must be filled from a fixed set of $\displaystyle \displaystyle p$ items.

    The first of these $\displaystyle \displaystyle p$ items can go in any one of the $\displaystyle \displaystyle r$ boxes; the next in any one of the remaining $\displaystyle \displaystyle (r-1)$ boxes; and so on. The number of ways of filling the first $\displaystyle \displaystyle p$ boxes from the fixed set is therefore
    $\displaystyle \displaystyle r(r-1)(r-2)...(r-p+1)={^rP}_p$
    There are now $\displaystyle \displaystyle (r-p)$ remaining boxes to be filled with items to be chosen from the remaining $\displaystyle \displaystyle (n-p)$. This can clearly be done in $\displaystyle \displaystyle {^{(n-p)}P}_{(r-p)}$ ways.

    So I reckon the total number of ways is
    $\displaystyle \displaystyle {^rP}_p\times{^{(n-p)}P}_{(r-p)}$
    Grandad
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  3. #3
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    That statement is really difficult to read and understand.

    Let me guess. Let’s say we have ten boys among which there are three brothers.
    How many ways to from a line of these ten so that the three bothers are together?
    The answer is $\displaystyle (3!)(8!)$. There are seven other boys and a group of three, making eight ‘distinct’ items to arrange. But one of those can be further arranged in $\displaystyle 3!$ ways.

    Is that what this question is about?
    If not, then please try to make it clear.
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  4. #4
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    Hello everyone

    I don't see any particular ambiguity in the way the question is phrased. I have assumed that it means:
    How many ways are there of choosing and arranging $\displaystyle \displaystyle r$ items from a set of $\displaystyle \displaystyle n$, if a certain group of $\displaystyle \displaystyle p$ items must always be included in the arrangement?
    The question Plato has answered is quite different. He has interpreted it as:
    How many ways are there of arranging $\displaystyle \displaystyle n$ items, if a certain group of $\displaystyle \displaystyle p$ items must be placed together in a block within the arrangement?
    There doesn't seem any room for the variable $\displaystyle \displaystyle r$ within this latter interpretation.

    Grandad
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    thank you so much,grandad and plato.
    yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.
    but i have the formula correctly copied.i have checked the language and the formula again...
    but i am having a problem with (r permutation p)which means i am finding the no of arrengements of r objects,taken p at a time. it is giving me a strange feeling that i am not fixing the p things in all arrengements.
    say i have the letters a,b,c,d,e,f of which i am asked to find the no of 4 letter words i can form ,where each word must have c and e.(ie. n=6,r=4, ,p=2)
    now if i only think about the letters other than c and e, i have 2 places to be filled from 4letters -a,b,d,f =4P2
    now to put c and e ,i have 3 blank spaces around the 2 words(other than c and e)chosen already by the above process.
    now c has 3 options and d also has 3 options, (because they can sit together),so 3*3*4P2?
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    Quote Originally Posted by mathlove View Post
    yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.

    giving me a strange feeling that i am not fixing the p things in all arrangements.
    There are $\displaystyle \displaystyle\binom{N-p}{r-p}=\displaystyle\frac{(N-p)!}{(r-p)!(N-r)!}$ to choose the $\displaystyle r$ items among which the specific $\displaystyle p$ items must be.
    Then there are $\displaystyle r!$ ways to arrange those.

    However, I am given pause when reading you phrase, “not fixing the p things in all arrangements”
    What does that mean? Is there a fixed place for the $\displaystyle p$ items?
    Last edited by Plato; Jul 2nd 2010 at 09:18 AM.
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  7. #7
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    Hello mathlove
    Quote Originally Posted by mathlove View Post
    thank you so much,grandad and plato.
    yes, actually it means that i always need the p fixed things to remain in each of my arrengement,but not necessarily together.
    but i have the formula correctly copied.i have checked the language and the formula again...
    but i am having a problem with (r permutation p)which means i am finding the no of arrengements of r objects,taken p at a time. it is giving me a strange feeling that i am not fixing the p things in all arrengements.
    say i have the letters a,b,c,d,e,f of which i am asked to find the no of 4 letter words i can form ,where each word must have c and e.(ie. n=6,r=4, ,p=2)
    now if i only think about the letters other than c and e, i have 2 places to be filled from 4letters -a,b,d,f =4P2
    now to put c and e ,i have 3 blank spaces around the 2 words(other than c and e)chosen already by the above process.
    now c has 3 options and d also has 3 options, (because they can sit together),so 3*3*4P2?
    No. You are assuming here that the other two letters, say a and b are not adjacent to one another.

    You must position the letters c and e first, choosing 2 empty 'boxes' from 4, and then arranging the letters within these boxes. This can be done in $\displaystyle {^4P}_2$ ways. The remaining two empty spaces can then be filled from the remaining 4 letters in $\displaystyle {^4P}_2$ ways.

    Total: $\displaystyle {^4P}_2\times{^4P}_2$ ways.

    Grandad
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    thank you,grandad. i have understood my mistake.thanks again.
    and i am facing some more problems in combinatorics,will i put them in this section'discrete maths'??
    i found that it was moved there.
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    Quote Originally Posted by mathlove View Post
    thank you,grandad. i have understood my mistake.thanks again.
    and i am facing some more problems in combinatorics,will i put them in this section'discrete maths'??
    i found that it was moved there.
    Yep, Discrete Mathematics, Set Theory and Logic is the appropriate place.
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