具体数学-第13课（组合数各种性质）

具体数学-第13课（组合数各种性质）

原文链接：
具体数学-第13课 - WeiYang Blog
首先庆祝我自己顺利毕业了，忙完了毕业论文答辩一直在浪，所以上周的具体数学没有更新，现在补更一下，大家见谅。

首先这节课讲的基本都是组合数的相关性质，而且特别多，所以我就不在这里详细证明了，如果你们对某一个性质感兴趣，可以自己证明去。

性质1

首先将组合数推广到负数域，也就是底数为负数的情况：
$left( {egin{array}{*{20}{c}}r\kend{array}} ight) = {( - 1)^k}left( {egin{array}{*{20}{c}}{k - r - 1}\kend{array}} ight)$
证明可以从下降阶乘幂的定义直接得到。

性质2

由于
$left( {egin{array}{*{20}{c}}{m + n}\mend{array}} ight) = left( {egin{array}{*{20}{c}}{m + n}\nend{array}} ight)$
所以由性质1可得
${( - 1)^m}left( {egin{array}{*{20}{c}}{ - n - 1}\mend{array}} ight) = {( - 1)^n}left( {egin{array}{*{20}{c}}{ - m - 1}\nend{array}} ight)$

性质3

$sumlimits_{k le m} {left( {egin{array}{*{20}{c}}r\kend{array}} ight){ {( - 1)}^k}} = {( - 1)^m}left( {egin{array}{*{20}{c}}{r - 1}\mend{array}} ight)$
这就说明了杨辉三角同一行的前面若干项交错和是可以求得的，但是它们的直接和是无法求出的。

性质4

$sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + r}\kend{array}} ight){x^k}{y^{m - k}} = sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{ - r}\kend{array}} ight){ {( - x)}^k}{ {(x + y)}^{m - k}}} }$
证明可以通过令
${S_m} = sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + r}\kend{array}} ight){x^k}{y^{m - k}}} = sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + r - 1}\kend{array}} ight){x^k}{y^{m - k}}} + sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + r - 1}\{k - 1}end{array}} ight){x^k}{y^{m - k}}}$
将左边表示成递归式的形式，同理如果右边可以表示成相同的递归式，那么左右就相等了。

性质4看起来特别复杂，那么它有什么用呢？如果令 $x$ 和 $y$ 等于不同的值，那么就可以得到许多不同的恒等式。

性质5

令 $x = - 1,y = 1$ 可以得到
$sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + r}\kend{array}} ight){ {( - 1)}^k}} = left( {egin{array}{*{20}{c}}{ - r}\mend{array}} ight)$
这其实就是性质3的特例。

性质6

令 $x = y = 1,r = m + 1$ 可以得到
$sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{2m + 1}\kend{array}} ight)} = sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + k}\kend{array}} ight){2^{m - k}}}$
左边就是杨辉三角一行中左边一半的和，所以可以得到
$sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{m + k}\kend{array}} ight){2^{ - k}}} { m{ = }}{2^m}$

性质7

$left( {egin{array}{*{20}{c}}r\mend{array}} ight)left( {egin{array}{*{20}{c}}m\kend{array}} ight) = left( {egin{array}{*{20}{c}}r\kend{array}} ight)left( {egin{array}{*{20}{c}}{r - k}\{m - k}end{array}} ight)$
这个公式可以形象理解为，从 $r$ 个物品中取 $m$ 个，再从这 $m$ 个中取 $k$ 个的方法数等于从 $r$ 个物品中取 $k$ 个，再从剩下的 $r-k$ 个中取 $m-k$ 个的方法数。证明的话直接用定义可证。

性质8

之前介绍了二项式系数，那么可以推广到任意 $m$ 个未知数，它的展开式为
${({x_1} + {x_2} + cdots + {x_m})^n} = sumlimits_{scriptstyle0 le {a_1},{a_2}, cdots ,{a_m} le natopscriptstyle{a_1} + {a_2} + cdots + {a_m} = n} {left( {egin{array}{*{20}{c}}{ {a_1} + {a_2} + cdots + {a_m}}\{ {a_1},{a_2}, cdots ,{a_m}}end{array}} ight)} {x_1}^{ {a_1}}{x_2}^{ {a_2}} cdots {x_m}^{ {a_m}}$
其中
$left( {egin{array}{*{20}{c}}{ {a_1} + {a_2} + cdots + {a_m}}\{ {a_1},{a_2}, cdots ,{a_m}}end{array}} ight) = left( {egin{array}{*{20}{c}}{ {a_1} + {a_2} + cdots + {a_m}}\{ {a_2} + cdots + {a_m}}end{array}} ight) cdots left( {egin{array}{*{20}{c}}{ {a_{m - 1}} + {a_m}}\{ {a_m}}end{array}} ight)$

性质9

范德蒙德卷积式：
$sumlimits_k {left( {egin{array}{*{20}{c}}r\{m + k}end{array}} ight)} left( {egin{array}{*{20}{c}}s\{n - k}end{array}} ight) = left( {egin{array}{*{20}{c}}{r + s}\{m + n}end{array}} ight)$
很多公式都可以通过替换其中的一些变量推导得到：
$egin{array}{l}sumlimits_k {left( {egin{array}{*{20}{c}}l\{m + k}end{array}} ight)} left( {egin{array}{*{20}{c}}s\{n + k}end{array}} ight) = left( {egin{array}{*{20}{c}}{l + s}\{l - m + n}end{array}} ight)\sumlimits_k {left( {egin{array}{*{20}{c}}l\{m + k}end{array}} ight)} left( {egin{array}{*{20}{c}}{s + k}\nend{array}} ight){( - 1)^k} = {( - 1)^{l + m}}left( {egin{array}{*{20}{c}}{s - m}\{n - l}end{array}} ight)\sumlimits_{k le l} {left( {egin{array}{*{20}{c}}{l - k}\mend{array}} ight)} left( {egin{array}{*{20}{c}}s\{k - n}end{array}} ight){( - 1)^k} = {( - 1)^{l + m}}left( {egin{array}{*{20}{c}}{s - m - 1}\{l - m - n}end{array}} ight)\sumlimits_{0 le k le l} {left( {egin{array}{*{20}{c}}{l - k}\mend{array}} ight)} left( {egin{array}{*{20}{c}}{q + k}\nend{array}} ight) = left( {egin{array}{*{20}{c}}{l + q + 1}\{m + n + 1}end{array}} ight)end{array}$

例题1

最后详细求解一道组合题，其他的题目就不介绍了，可以去看具体数学英文版第173页。

求下面式子的闭形式解：
$sumlimits_{k = 0}^m {left( {egin{array}{*{20}{c}}m\kend{array}} ight)/left( {egin{array}{*{20}{c}}n\kend{array}} ight)} ,n ge m ge 0$

根据性质7，可以得到
$left( {egin{array}{*{20}{c}}m\kend{array}} ight)/left( {egin{array}{*{20}{c}}n\kend{array}} ight) = left( {egin{array}{*{20}{c}}{n - k}\{m - k}end{array}} ight)/left( {egin{array}{*{20}{c}}n\mend{array}} ight)$
所以
$sumlimits_{k = 0}^m {left( {egin{array}{*{20}{c}}m\kend{array}} ight)/left( {egin{array}{*{20}{c}}n\kend{array}} ight)} = sumlimits_{k = 0}^m {left( {egin{array}{*{20}{c}}{n - k}\{m - k}end{array}} ight)/left( {egin{array}{*{20}{c}}n\mend{array}} ight)}$
而
$egin{array}{l}sumlimits_{k ge 0} {left( {egin{array}{*{20}{c}}{n - k}\{m - k}end{array}} ight)} = sumlimits_{m - k ge 0} {left( {egin{array}{*{20}{c}}{n - (m - k)}\{m - (m - k)}end{array}} ight)} \ = sumlimits_{k le m} {left( {egin{array}{*{20}{c}}{n - m + k}\kend{array}} ight)} \ = left( {egin{array}{*{20}{c}}{(n - m) + m + 1}\mend{array}} ight)\ = left( {egin{array}{*{20}{c}}{n + 1}\mend{array}} ight)end{array}$
所以
$sumlimits_{k = 0}^m {left( {egin{array}{*{20}{c}}m\kend{array}} ight)/left( {egin{array}{*{20}{c}}n\kend{array}} ight)} = left( {egin{array}{*{20}{c}}{n + 1}\mend{array}} ight)/left( {egin{array}{*{20}{c}}n\mend{array}} ight) = frac{ {n + 1}}{ {n + 1 - m}}$
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原文地址：https://www.cnblogs.com/godweiyang/p/12203921.html

具体数学-第13课（组合数各种性质）

原文链接：

性质1

性质2

性质3

性质4

性质5

性质6

性质7

性质8

性质9

例题1