Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
搜尋書籍內容
第 1 到 3 筆結果,共 53 筆
第 126 頁
... Consequently , the coefficient of x " is ( −1 ) m | 0 for m = n ( compare with part b ) . 57g . The sum to be evaluated equals the coefficient of x * in the polynomial ( 1 + x ) " + ( 1 + x ) ” + 1 + ( 1 + x ) n + 2 + · · · + ( 1 + x ) ...
... Consequently , the coefficient of x " is ( −1 ) m | 0 for m = n ( compare with part b ) . 57g . The sum to be evaluated equals the coefficient of x * in the polynomial ( 1 + x ) " + ( 1 + x ) ” + 1 + ( 1 + x ) n + 2 + · · · + ( 1 + x ) ...
第 158 頁
... consequently , also the required probability p1 ) equals zero : if the total number of white balls is odd , at least one of the people must draw one white ball and one black ball . Hence we have only to consider the case of even n = 2k ...
... consequently , also the required probability p1 ) equals zero : if the total number of white balls is odd , at least one of the people must draw one white ball and one black ball . Hence we have only to consider the case of even n = 2k ...
第 182 頁
... Consequently , for n ≥ 2m the number of shortest paths from A to An + m which have vertices on the line Ĩ , is 2N Ao An + m Now it is easy to answer the question raised in the problem . For n < 2m there are no favorable outcomes at all ...
... Consequently , for n ≥ 2m the number of shortest paths from A to An + m which have vertices on the line Ĩ , is 2N Ao An + m Now it is easy to answer the question raised in the problem . For n < 2m there are no favorable outcomes at all ...
其他版本 - 查看全部
常見字詞
A₁ A₂ An+m arrangements b₁ B₂ binomial coefficients binomial theorem bishops black squares C₁ chessboard chord circle coefficient color column compute the number Consequently consider corresponding customers denote determine the number diagonals digits dihedral angle divided divisible draw equally likely possible equation equidistant equivalence classes exactly example experiment favorable outcomes follows formula G₁ given Hence inclusion and exclusion intersection k-gons knights length mathematical induction maximum number n-gon number of different number of favorable number of paths number of shortest obtain pairs partition passengers plane polygons positive integers possible outcomes Pr{E probability theory problem 54 prove queens rectangle relatively prime remaining required probability rooks S₁ segment selected at random sequence shortest paths side solution to problem solved sphere square controlled Suppose T₂ total number triangle unfavorable values vertex vertices