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 23 筆
第 61 頁
... coefficient of x18 is zero ) . = The number 17 can be written in exactly one way as a sum of 5's and 7's : 175 +5 +7 . Consequently , the coefficient of x17 is equal to the number of terms obtained by selecting x5 from two of the ...
... coefficient of x18 is zero ) . = The number 17 can be written in exactly one way as a sum of 5's and 7's : 175 +5 +7 . Consequently , the coefficient of x17 is equal to the number of terms obtained by selecting x5 from two of the ...
第 126 頁
... coefficient of x " in the polynomial x " ( 1 − x ) " + x ^ −1 ( 1 − x ) " + ··· + xn − m ( 1 − x ) ... 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 ...
... coefficient of x " in the polynomial x " ( 1 − x ) " + x ^ −1 ( 1 − x ) " + ··· + xn − m ( 1 − x ) ... 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 ...
第 127 頁
... coefficient of x2 " in the expression [ 1 − x2n + 1 ( 1 − x ) 2n + 1 ] = - 1 - x + x2 2 [ 1 − x2n + 1 ( 1 − x ) 2n + 1 ] 1 + x 1 + x3 By virtue of the formula for the sum ... coefficient of x V. Problems on the binomial coefficients 127.
... coefficient of x2 " in the expression [ 1 − x2n + 1 ( 1 − x ) 2n + 1 ] = - 1 - x + x2 2 [ 1 − x2n + 1 ( 1 − x ) 2n + 1 ] 1 + x 1 + x3 By virtue of the formula for the sum ... coefficient of x V. Problems on the binomial coefficients 127.
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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