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Math Help - Divinding 8 teachers up among 4 schools.

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    Divinding 8 teachers up among 4 schools.

    28. If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers?

    OK, so it's pretty clear that if x_i is the number of teachers allocated to one of the four schools then x_1+x_2+x_3+x_4=8

    Comparing this too the chapter text leads me to the following, \binom{n+r-1}{r-1}, which is the number of distinct nonnegative integer-valued vectors (x_1, x_2, . . . , x_r) satisfying the equation x_1 + x_2 +...+ x_r=n

    So my first question is, since I need x_i=0, to be a valid entry does this formula cover that case? (I'm thinking it does because it mentions nonnegative integer-valued vectors.)

    Secondly, taking n=8 and r=4, I've got \binom{8+4-1}{4-1}=\frac{11!}{8!3!}=165 as my current working solution. However, the text is talking about indistinguishable objects here and I hardly think teachers are indistinguishable. Also, the listed answer is 65,536. So I think I need to permute the teachers now somehow? Or is this all wrong?
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by bkbowser View Post
    28. If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers?

    OK, so it's pretty clear that if x_i is the number of teachers allocated to one of the four schools then x_1+x_2+x_3+x_4=8

    Comparing this too the chapter text leads me to the following, \binom{n+r-1}{r-1}, which is the number of distinct nonnegative integer-valued vectors (x_1, x_2, . . . , x_r) satisfying the equation x_1 + x_2 +...+ x_r=n

    So my first question is, since I need x_i=0, to be a valid entry does this formula cover that case? (I'm thinking it does because it mentions nonnegative integer-valued vectors.)

    Secondly, taking n=8 and r=4, I've got \binom{8+4-1}{4-1}=\frac{11!}{8!3!}=165 as my current working solution. However, the text is talking about indistinguishable objects here and I hardly think teachers are indistinguishable. Also, the listed answer is 65,536. So I think I need to permute the teachers now somehow? Or is this all wrong?
    for your first question the simple example of n=1, r=2 shows that the case of x_i=0 is covered by this formula.

    for your second question, ignoring what you've posted, think about it this way. Each teacher can be assigned to 1 of 4 schools. Think of it as giving them a base 4 digit. there are 8 teachers so the total number of variations is 4^8=65536.
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by bkbowser View Post
    28. If 8 new teachers are to be divided among 4 schools, how many divisions are possible? What if each school must receive 2 teachers?
    However, the text is talking about ndistinguishable objects here and I hardly think teachers are indistinguishable. Also, the listed answer is 65,536. So I think I need to permute the teachers now somehow? Or is this all wrong?
    First I assume that there are two distinct questions here. 1) there are no restrictions. 2) each school gets two teachers.

    1) There are m^n functions from a set of n elements to a set of m elements.

    2) How many ways can we rearrange the string AABBCCDD~?
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by romsek View Post
    for your first question the simple example of n=1, r=2 shows that the case of x_i=0 is covered by this formula.

    for your second question, ignoring what you've posted, think about it this way. Each teacher can be assigned to 1 of 4 schools. Think of it as giving them a base 4 digit. there are 8 teachers so the total number of variations is 4^8=65536.
    With n=1,r=2 zero would have to be an option since we only care about integer solutions. \binom{1+2-1}{2-1}=\binom{2}{1}=2. OK good.

    Assigning each teacher a number out of a base 4 system seems to work well, but can I use the

    Quote Originally Posted by Plato View Post
    First I assume that there are two distinct questions here. 1) there are no restrictions. 2) each school gets two teachers.

    1) There are m^n functions from a set of n elements to a set of m elements.

    2) How many ways can we rearrange the string AABBCCDD~?
    Sorry, yes there are two questions in the original, I didn't actually get around to trying to solve the second as I wasn't finished with the first.

    1) looks very similar to romsek's suggestion. Is there no way to use this proposition from the text to solve part 1? \binom{n+r-1}{r-1}, which is the number of distinct nonnegative integer-valued vectors (x_1, x_2, . . . , x_r) satisfying the equation x_1 + x_2 +...+ x_r=n.

    For 2) there are essentially 8 slots and 8 options for the first slot. The options decrease by one with every slot so there are 8! ways? No this can't be right.
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by bkbowser View Post
    1) looks very similar to romsek's suggestion. Is there no way to use this proposition from the text to solve part 1? \binom{n+r-1}{r-1}, which is the number of distinct nonnegative integer-valued vectors (x_1, x_2, . . . , x_r) satisfying the equation x_1 + x_2 +...+ x_r=n.

    For 2) there are essentially 8 slots and 8 options for the first slot. The options decrease by one with every slot so there are 8! ways? No this can't be right.
    NO! 1) is 4^8. The number of functions from eight teachers to four schools.

    2) the answer is \frac{8!}{(2)^4}.
    Think of the schools as \{A,B,C,D\} Make a list of the eight teachers.
    Arrange the string AABBCCDD, one next to each name.
    Every rearrangement of that string is a possible way to assign eight teachers to four schools with two to each school.
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by Plato View Post
    NO! 1) is 4^8. The number of functions from eight teachers to four schools.

    2) the answer is \frac{8!}{(2)^4}.
    Think of the schools as \{A,B,C,D\} Make a list of the eight teachers.
    Arrange the string AABBCCDD, one next to each name.
    Every rearrangement of that string is a possible way to assign eight teachers to four schools with two to each school.
    I don't fully follow but I think I can make sense of it with the text as follows; So this is a Multinomial Coefficient problem where each of the r=4 distinct subgroups is 2, which the proof in the text shows is equal to \frac{n!}{n_1!n_2!n_3!n_4!}, and this is equal to \frac{8!}{2!2!22!}=\frac{8!}{2^4}=2520.

    Does this look good?
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by bkbowser View Post
    and this is equal to \frac{8!}{2!2!22!}=\frac{8!}{2^4}=2520.
    Does this look good?
    Yes, that is correct.

    But the other model is very useful.

    Do you understand how many ways you can rearrange the word MISSISSIPPI~?
    Well \frac{11!}{(4!)^2(2!)}.
    That is also how many ways that a collection of eleven people can be divided into four different cells where two cells contain four people, another contains two and the fourth contains one.
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by Plato View Post
    Do you understand how many ways you can rearrange the word MISSISSIPPI~?
    Well \frac{11!}{(4!)^2(2!)}.
    That is also how many ways that a collection of eleven people can be divided into four different cells where two cells contain four people, another contains two and the fourth contains one.
    I'm not sure I understand the model.

    11! is the permutations of an 11 place word.

    I know you have to take out the number of duplicates, which I presume is the denominator, I'm just not sure how.
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    Re: Divinding 8 teachers up among 4 schools.

    Quote Originally Posted by bkbowser View Post
    I'm not sure I understand the model.
    11! is the permutations of an 11 place word.
    I know you have to take out the number of duplicates, which I presume is the denominator, I'm just not sure how.
    Put subscripts on any repeated letters: MI_1S_1S_2I_2S_3S_3I_3P_1P_2I_4.
    Now we have eleven different letters. We can rearrange that string in 11! ways.

    Those subscripted I's can appear in 4! ways without changing relative places in the string.
    Those subscripted S's can appear in 4! ways without changing relative places in the string.
    Those subscripted P's can appear in 2! ways without changing relative places in the string.

    So if we drop all the subscripts, we have to divide by those numbers to get the actual count.
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