# Mathematical Questions and Solutions Volume 53

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This historic book may have numerous typos and missing text. Purchasers can usually download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1890 edition. Excerpt: ...the order of the drawing, and show that this chance approaches to el as the numbers k and increase." The number of all possible cases is 1. 2... M, say P . The number of cases the first ball drawn is B1 is P _i. The number of cases the second ball drawn is Bs (without the first having been B, ) is P _i-P -2. The number of cases the third ball drawn is B3 (without the first having been Bi or the second B2) is Pn-I--2P _ 2 + P-3, &c. Thus we find P -l, P -l-P -2, P, .1-2P .2 + P .3, P _, -3P -2 + 3P _3-P - and, with the notation of binomial coefficients, at last (Y)p--The sum of the first k lines is or p( 1 (-!) 1 1 (-l)(-2) 1 "t 1.2 (-1) 1.2.3 (-l)(-2) "-\$' So the chance W, in question is given by the equation W-1-1 1 1 1 k(k-l)(k-2) '" 1 1.2 (-!) 1.2.3 (-l)(-2)-"' which proves that W, approaches to l/ when k and n increase. For an analogous problem, Professor Schoute refers to Wiskundige Opgaven, tome iv., page 96, problem 34. 10244. (F. R. J. Hrrvey.)--Prove that the inverses of a point P with respect to four mutually orthogonal circles, invert from P and from three other points into mutually orthocentric points. Solution by Professors Murhopadhyay, Beyens, and others. It is known that a circle and a pair of inverse points with respect to that circle invert from any arbitrary point into a circle and a pair of inverse points with respect to this latter circle. (casey's Sequel, M. Ex. 60.) Let Plt P2, P3, P4 be the inverses of the point P with respect to the four mutually orthogonal circles A, B, C, D. Invert the system from...