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Math Help - 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 r empty boxes to fill, and \displaystyle p of them must be filled from a fixed set of \displaystyle p items.

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

    So I reckon the total number of ways is
    \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 (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 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 r items from a set of \displaystyle n, if a certain group of \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 n items, if a certain group of \displaystyle p items must be placed together in a block within the arrangement?
    There doesn't seem any room for the variable \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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  6. #6
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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\binom{N-p}{r-p}=\displaystyle\frac{(N-p)!}{(r-p)!(N-r)!} to choose the r items among which the specific p items must be.
    Then there are 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 p items?
    Last edited by Plato; July 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 {^4P}_2 ways. The remaining two empty spaces can then be filled from the remaining 4 letters in {^4P}_2 ways.

    Total: {^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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