Challenging Mathematical Problems with Elementary Solutions: Combinatorial analysis and probability theoryHolden-Day, 1964 |
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第 1 到 3 筆結果,共 39 筆
第 48 頁
... remaining colors , say red , and turn the cube so that the red face is the back face . This can always be done by rotating the cube about the vertical axis through its center ( which leaves the blue face on top and the green face on the ...
... remaining colors , say red , and turn the cube so that the red face is the back face . This can always be done by rotating the cube about the vertical axis through its center ( which leaves the blue face on top and the green face on the ...
第 147 頁
... remaining car . Hence there are 3.2 ( 2 ) ( 3 ) = 7560 favorable outcomes , and the required probability is 7560/39 ≈ 0.384 . ( This probability is 4 times as great as that of the arrangement in part b . ) 70. The experiment considered ...
... remaining car . Hence there are 3.2 ( 2 ) ( 3 ) = 7560 favorable outcomes , and the required probability is 7560/39 ≈ 0.384 . ( This probability is 4 times as great as that of the arrangement in part b . ) 70. The experiment considered ...
第 158 頁
... remaining k people can likewise each pick a pair of black balls in ( 2k ) ! / 2 * different ways . Combining these ... remaining m - 1 addresses can be written on the second envelope ; then any of the remaining m 2 addresses on the third ...
... remaining k people can likewise each pick a pair of black balls in ( 2k ) ! / 2 * different ways . Combining these ... remaining m - 1 addresses can be written on the second envelope ; then any of the remaining m 2 addresses on the third ...
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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