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Hint:Assume a chess board of \[8\times 8\]. Find the different ways of dimensions possible to determine the number of rectangles.There are 9 horizontal lines and 9 vertical lines. A rectangle will be formed by two vertical and two horizontal lines.And two horizontal lines can be selected in ${}^{9}{{C}_{2}}$ ways.Similarly, number of ways to select 2 lines out of 9 vertical lines is also ${}^{9}{{C}_{2}}$.Thus formulate and find the number of rectangles.
Complete step-by-step answer:
Let us assume that it is a \[8\times 8\] chessboard. To form a rectangle in the chess board, it can have the following dimension.
\[\begin{align}
& 1\times 1,1\times 2,1\times 3,1\times 4,1\times 5,1\times 6,1\times 7,1\times 8 \\
& 2\times 2,2\times 3,2\times 4,2\times 5,2\times 6,2\times 7,2\times 8 \\
& 3\times 3,3\times 4,3\times 5,3\times 6,3\times 7,3\times 8 \\
& 4\times 4,4\times 5,4\times 6,4\times 7,4\times 8 \\
& 5\times 5,5\times 6,5\times 7,5\times 8 \\
& 6\times 6,6\times 7,6\times 8 \\
& 7\times 7,7\times 8 \\
& 8\times 8 \\
\end{align}\]
Thus we got 36 ways of how to form the dimension of the rectangle in a chess board. To form a rectangle, we need 4 lines i.e. 2 sets of parallel lines.
We can generalize the total number of rectangles in a \[n\times n\] chess board to \[{}^{n+1}{{C}_{2}}\times {}^{n+1}{{C}_{2}}\].
Thus we know that \[n\times n\] is equal to \[8\times 8\].
Thus applying n in the above formula,
Number of rectangles \[={}^{8+1}{{C}_{2}}\times {}^{8+1}{{C}_{2}}={}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}\]
They are of the form, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[{}^{9}{{C}_{2}}=\dfrac{9!}{2!\left( 9-2 \right)!}=\dfrac{9!}{7!2!}=\dfrac{9\times 8\times 7!}{7!\times 2\times 1}=\dfrac{9\times 8}{2}=9\times 4=36\]
\[\therefore \] Number of rectangles \[={}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}=36\times 36=1296\].
\[\therefore \] The number of rectangles that can be formed on a chess board = 1296.
Note: If the chess board is of the dimension \[m*n\], then the total number of rectangles on the \[m*n\] board will be, \[{}^{m+1}{{C}_{2}}\times {}^{n+1}{{C}_{2}}\], where a rectangle can be formed by selecting 2 lines from (m + 1) and 2 lines from (n + 1).
Complete step-by-step answer:
Let us assume that it is a \[8\times 8\] chessboard. To form a rectangle in the chess board, it can have the following dimension.
\[\begin{align}
& 1\times 1,1\times 2,1\times 3,1\times 4,1\times 5,1\times 6,1\times 7,1\times 8 \\
& 2\times 2,2\times 3,2\times 4,2\times 5,2\times 6,2\times 7,2\times 8 \\
& 3\times 3,3\times 4,3\times 5,3\times 6,3\times 7,3\times 8 \\
& 4\times 4,4\times 5,4\times 6,4\times 7,4\times 8 \\
& 5\times 5,5\times 6,5\times 7,5\times 8 \\
& 6\times 6,6\times 7,6\times 8 \\
& 7\times 7,7\times 8 \\
& 8\times 8 \\
\end{align}\]
Thus we got 36 ways of how to form the dimension of the rectangle in a chess board. To form a rectangle, we need 4 lines i.e. 2 sets of parallel lines.
We can generalize the total number of rectangles in a \[n\times n\] chess board to \[{}^{n+1}{{C}_{2}}\times {}^{n+1}{{C}_{2}}\].
Thus we know that \[n\times n\] is equal to \[8\times 8\].
Thus applying n in the above formula,
Number of rectangles \[={}^{8+1}{{C}_{2}}\times {}^{8+1}{{C}_{2}}={}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}\]
They are of the form, \[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\].
\[{}^{9}{{C}_{2}}=\dfrac{9!}{2!\left( 9-2 \right)!}=\dfrac{9!}{7!2!}=\dfrac{9\times 8\times 7!}{7!\times 2\times 1}=\dfrac{9\times 8}{2}=9\times 4=36\]
\[\therefore \] Number of rectangles \[={}^{9}{{C}_{2}}\times {}^{9}{{C}_{2}}=36\times 36=1296\].
\[\therefore \] The number of rectangles that can be formed on a chess board = 1296.
Note: If the chess board is of the dimension \[m*n\], then the total number of rectangles on the \[m*n\] board will be, \[{}^{m+1}{{C}_{2}}\times {}^{n+1}{{C}_{2}}\], where a rectangle can be formed by selecting 2 lines from (m + 1) and 2 lines from (n + 1).
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