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Thus bNik) = 1 if ft is odd and 7^ (ft + 1) = 1, i.e., IN(k + 1 ) = 0. By (8), we have (6). Finally, for even ft < n — 1 and N an odd multiple of (ft + l)(ft + 2) • • • n, the mark ft + 1 will have made an odd number of passages through the smaller marks and will be at the left of an even permutation of the marks from 1 to ft. Thus the permutation of the marks from 1 to ft 41 is also even since ft is even and so Iw-i(k + 2) = 0 with ft 42 at the extreme left. The permutation on the marks 1, 2, • • -, ft 42 is then odd, so that ft 43 is at the right of these marks with 7jv_i(fc + 3) = 1, as are the marks larger than ft 43. The permutation PN-i is of the type

Determinants by means of solutions of linear equations.n63