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  1. #1
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    Exclamation help!

    Q1-Total number of squares of any size(side being natural nos.) in a rectangle of  \[m\times n(m<n),(m,n\in N)\]
    Q2-In how many ways can 15 boys and 3 girls can sit in a row such that between the girls at most two boys sit?
    all the persons are distinguishable and their order must be considred.
    Last edited by Navesh; June 23rd 2006 at 11:00 PM.
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  2. #2
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    Hello, Navesh!

    These problems raise even more questions . . .

    Q1) Total number of squares of any size (side being natural nos.)
    . . . in an m \times n rectangle . . . m < n,\;m,n \in  N
    Are we allowed to use the Greatest Integer Function?
    . . [x] = the greatest integer less than or equal to x.
    Basically, it is a "round down" function.


    Then we count the number of squares of various sizes . . .

    1\times1:\;\;m\cdot n

    2\times2:\;\;\left[\frac{m}{2}\right]\cdot\left[\frac{n}{2}\right]

    3\times3:\;\;\left[\frac{m}{3}\right]\cdot\left[\frac{n}{3}\right]

    4\times4:\;\;\left[\frac{m}{4}\right]\cdot\left[\frac{n}{4}\right]

    . . \vdots . . . . . . . \vdots

    m\times m:\;\;1\cdot\left[\frac{n}{m}\right]

    Then add these numbers.


    Q2) In how many ways can 15 boys and 3 girls can sit in a row
    . . . such that between the girls at most two sit?
    (a) If the 18 children are distinguishable (they have different names),
    . . then their order must be considered.

    (b)If the 15 boys and 3 girls are indistinguishable (arrange 15 blue marbles
    . . and 3 red marbles in a row), the problem is still quite difficult.
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  3. #3
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    Quote Originally Posted by Navesh
    Q1-Total number of squares of any size(side being natural nos.) in a rectangle of  \[m\times n(m<n),(m,n\in N)\]
    Q2-In how many ways can 15 boys and 3 girls can sit in a row such that between the girls at most two boys sit?
    all the persons are distinguishable and their order must be considred.
    The answer given in my textbook is \[\sum (m-r)(n-r), r=0,1,2,\ldots ,m-1\] .Why is it so? I think you are missing some of the squares.
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  4. #4
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    Hello, Navesh!

    The answer given in my textbook is: \[\sum^{m-1}_{r=0} (m-r)(n-r).

    Of course, it is! . . . I looked at it from the worst direction . . . *blush*
    I've solved this type of problem before, but totally forgot the approach.

    We have an m \times n "chessboard" with m rows and n columns.

    How many 1\!\times\!1 squares are there? .Let's call them \text{one-squares}.
    . . Place a \text{one-square} in the upper-left corner.
    . . There are mof them in the first row, and there are n such rows.
    Hence, there are: (m)(n)\text{ one-squares.}

    How many \text{two-squares} are there?
    . . Place a \text{two-square} in the upper-left corner.
    . . There are n-1 of them in the first row, and there are m-1 such rows.
    Hence, there are: (m-1)(n-1)\text{ two-squares.}

    How many \text{three-squares} are there?
    . . Place a \text{three-square} in the upper-left corner.
    . . There are n-2 of them in the first row, and there are m-2 such rows.
    Hence, there: (m-2)(n-2)\text{ three-squares.}
    . . . \vdots
    How many \text{m-squares} are there?
    . . Place an \text{m-square} in the upper-left corner.
    . . There are n-m+1 of them in the first row, and there is 1 such row.
    Hence, there are: 1(n-m+1)\;\text{m-squares.}


    Therefore, the total number of squares is:
    . . m\cdot n + (m-1)(n-1) + (m-2)(m-2) +  \hdots + 1(n-m+1)

    which can be written: . \sum^{m-1}_{r=0}(m - r)(n - r)
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  5. #5
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    Quote Originally Posted by Soroban
    Hello, Navesh!


    Of course, it is! . . . I looked at it from the worst direction . . . *blush*
    I've solved this type of problem before, but totally forgot the approach.

    We have an m \times n "chessboard" with m rows and n columns.

    How many 1\!\times\!1 squares are there? .Let's call them \text{one-squares}.
    . . Place a \text{one-square} in the upper-left corner.
    . . There are mof them in the first row, and there are n such rows.
    Hence, there are: (m)(n)\text{ one-squares.}

    How many \text{two-squares} are there?
    . . Place a \text{two-square} in the upper-left corner.
    . . There are n-1 of them in the first row, and there are m-1 such rows.
    Hence, there are: (m-1)(n-1)\text{ two-squares.}

    How many \text{three-squares} are there?
    . . Place a \text{three-square} in the upper-left corner.
    . . There are n-2 of them in the first row, and there are m-2 such rows.
    Hence, there: (m-2)(n-2)\text{ three-squares.}
    . . . \vdots
    How many \text{m-squares} are there?
    . . Place an \text{m-square} in the upper-left corner.
    . . There are n-m+1 of them in the first row, and there is 1 such row.
    Hence, there are: 1(n-m+1)\;\text{m-squares.}


    Therefore, the total number of squares is:
    . . m\cdot n + (m-1)(n-1) + (m-2)(m-2) +  \hdots + 1(n-m+1)

    which can be written: . \sum^{m-1}_{r=0}(m - r)(n - r)
    Thanks for help
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