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Michael Hardy
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find the probability that an n x n$n \times n$ determinant formed by taking the numbers 1$1, 2, ...\ldots, n^2n^2$ at random is odd

In the January 1963 (p.  94) issue of the American Mathematical Monthly, D.C.B. Marsh proved that for a 3 x 3$3 \times 3$ determinant formed from a random distribution of the integers 1, 2, ..., 9,$1, 2, \ldots, 9,$ the probability that it is odd is 4/7.$4/7.$ His proof was based on the expansion of the determinant into six terms of three products each and counting all possible ways the determinant could be odd. However, even with the 4 x 4$4 \times 4$ determinant, the expansion consists of 24$24$ terms of 4$4$ products each and it is not clear to me how to apply his method. If one looks at the 4 x 4$4 \times 4$ determinant itself and decides to construct it row by row, which must all be linearly independent, the first row can be anything except all even numbers. Thus the number of possible first rows is 16x15x14x13 minus to number of all possible even first rows which is 7x6x5x4.$7\times6\times5\times4.$ But the number of second rows is 121x11x10x9$121 \times 11 \times 10 \times 9$ minus the number of even rows, which alas is now indeterminate. Perhaps one can obtain a recursion formula for determinants of order n$n$ from those of order n-1$n-1$ but I am unable to find one.

find the probability that an n x n determinant formed by taking the numbers 1, 2, ..., n^2 at random is odd

In the January 1963 (p.94) issue of the American Mathematical Monthly, D.C.B. Marsh proved that for a 3 x 3 determinant formed from a random distribution of the integers 1, 2, ..., 9, the probability that it is odd is 4/7. His proof was based on the expansion of the determinant into six terms of three products each and counting all possible ways the determinant could be odd. However, even with the 4 x 4 determinant, the expansion consists of 24 terms of 4 products each and it is not clear to me how to apply his method. If one looks at the 4 x 4 determinant itself and decides to construct it row by row, which must all be linearly independent, the first row can be anything except all even numbers. Thus the number of possible first rows is 16x15x14x13 minus to number of all possible even first rows which is 7x6x5x4. But the number of second rows is 121x11x10x9 minus the number of even rows, which alas is now indeterminate. Perhaps one can obtain a recursion formula for determinants of order n from those of order n-1 but I am unable to find one.

find the probability that an $n \times n$ determinant formed by taking the numbers $1, 2, \ldots, n^2$ at random is odd

In the January 1963 (p.  94) issue of the American Mathematical Monthly, D.C.B. Marsh proved that for a $3 \times 3$ determinant formed from a random distribution of the integers $1, 2, \ldots, 9,$ the probability that it is odd is $4/7.$ His proof was based on the expansion of the determinant into six terms of three products each and counting all possible ways the determinant could be odd. However, even with the $4 \times 4$ determinant, the expansion consists of $24$ terms of $4$ products each and it is not clear to me how to apply his method. If one looks at the $4 \times 4$ determinant itself and decides to construct it row by row, which must all be linearly independent, the first row can be anything except all even numbers. Thus the number of possible first rows is 16x15x14x13 minus to number of all possible even first rows which is $7\times6\times5\times4.$ But the number of second rows is $121 \times 11 \times 10 \times 9$ minus the number of even rows, which alas is now indeterminate. Perhaps one can obtain a recursion formula for determinants of order $n$ from those of order $n-1$ but I am unable to find one.

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find the probability that an n x n determinant formed by taking the numbers 1, 2, ..., n^2 at random is odd

In the January 1963 (p.94) issue of the American Mathematical Monthly, D.C.B. Marsh proved that for a 3 x 3 determinant formed from a random distribution of the integers 1, 2, ..., 9, the probability that it is odd is 4/7. His proof was based on the expansion of the determinant into six terms of three products each and counting all possible ways the determinant could be odd. However, even with the 4 x 4 determinant, the expansion consists of 24 terms of 4 products each and it is not clear to me how to apply his method. If one looks at the 4 x 4 determinant itself and decides to construct it row by row, which must all be linearly independent, the first row can be anything except all even numbers. Thus the number of possible first rows is 16x15x14x13 minus to number of all possible even first rows which is 7x6x5x4. But the number of second rows is 121x11x10x9 minus the number of even rows, which alas is now indeterminate. Perhaps one can obtain a recursion formula for determinants of order n from those of order n-1 but I am unable to find one.