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[组合] 一道模拟考试中的计数问题

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aishuxue Posted at 2014-1-13 22:27:39 |Read mode
360截图20140113222642609.png

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乌贼 Posted at 2014-1-13 23:05:05
Last edited by 乌贼 at 2014-1-14 01:27:00这是什么题啊?由题意知序数组里$2$与$-2$的个数相差不能超过两个,再考虑每两个同号
捆绑一起后排序……还是不知道怎么作
若干个$a,b$排队有多少种排法?

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乌贼 Posted at 2014-1-13 23:19:45
回复 2# 乌贼
分$m$为奇偶两种情况:$1)$奇数时异号个数相差$2$个
                                $2)$偶数是异号个数相同
坐沙发

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其妙 Posted at 2014-1-13 23:51:31
回复 3# 乌贼
看到排列组合计数题脑壳就大,乌贼怎么不研究几何了啊?
isee搞了一个几何题,就等你去了!

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乌贼 Posted at 2014-1-14 00:02:05
回复 4# 其妙
最近脑袋进水了

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爪机专用 Posted at 2014-1-14 02:36:18
想是想到了,就是表达起来可能不太容易,画个折线图可能方便理解,时间关系先睡一觉,明天要是还没人写答案再说。。。
I am majia of kuing

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Tesla35 Posted at 2014-1-14 10:14:53
怎么不给出前几问,说不定有提示作用

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乌贼 Posted at 2014-1-14 11:32:06
Last edited by 乌贼 at 2014-1-14 17:21:00从后往前排(只考虑$2$)
$2,2,-2,-2\left\{\begin{aligned}2,2,-2,-2\left\{\begin{aligned}2,2,-2,-2\left\{\begin{aligned}2,2,-2,-2\\-2,-2,2,2\end{aligned}\right.\\-2,-2,2,2\left\{\begin{aligned}2,2,-2,-2\\-2,-2,2,2\end{aligned}\right.\end{aligned}\right.\\-2,-2,2,2\left\{\begin{aligned}2,2,-2,-2\left\{\begin{aligned}2,2,-2,-2\\-2,-2,2,2\end{aligned}\right.\\-2,-2,2,2\left\{\begin{aligned}2,2,-2,-2\\-2,-2,2,2\end{aligned}\right.\end{aligned}\right.\end{aligned}\right.\cdots$
$m$从$1$开始,每增加$2$,组合数就翻一倍。

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乌贼 Posted at 2014-1-14 12:11:39
回复 8# 乌贼
$m=2t,m=2t-1(t=1,2,3,\cdots)$的结果一样

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goft Posted at 2014-1-14 13:07:34
递推关系的感觉,先考察m=1,2,3,猜想归纳

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乌贼 Posted at 2014-1-14 13:11:11
人教群by冯加明
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乌贼 Posted at 2014-1-14 13:27:43
回复 11# 乌贼
当$m=2$时怎么有$10$组?

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乌贼 Posted at 2014-1-14 13:48:31
我怎么感觉把$m$分成奇偶数,令$m=2t$及$m=2t-1,(t=1,2,3,\cdots)$,
个数为  $2^t$
lz答案是多少?

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kuing Posted at 2014-1-14 16:46:14
想是想到了,就是表达起来可能不太容易,画个折线图可能方便理解,时间关系先睡一觉,明天要是还没人写答案 ...
爪机专用 发表于 2014-1-14 02:36

还是将昨晚想的完整过程写了一下,见附件(附带源码,用 xelatex 编译)。
真的不太会用数学语言来表达,所以后面都用了折线图表示,而且颇为啰嗦,希望能帮助理解。
$type

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$type

20140114yazhoujishu_kuing.tex

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乌贼 Posted at 2014-1-14 16:59:43
Last edited by 乌贼 at 2014-1-14 17:09:00回复 14# kuing
与11楼一样。
问题还是12楼,当$m=2$时只有$(2,2,-2,-2),(-2,-2,2,2)$两组啊!

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kuing Posted at 2014-1-14 17:25:02
回复  kuing
与11楼一样。
问题还是12楼,当$m=2$时只有$(2,2,-2,-2),(-2,-2,2,2)$两组啊! ...
乌贼 发表于 2014-1-14 16:59

结果一样,过程不一样……我还有递推法哩……[得意]
$m=2$ 时符合的有 $(2,-2,2,2)$, $(-2,2,2,2)$, $(2,2,-2,2)$, $(2,2,2,-2)$, $(2,2,-2,-2)$ 以及它们的反相数,共十个。

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乌贼 Posted at 2014-1-14 17:38:53
回复 16# kuing
那$\dfrac{a_{2k-1}}{a_{2k}}=-1$了

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kuing Posted at 2014-1-14 17:41:11
回复 17# 乌贼

有说完全不让 $a_{2k-1}/a_{2k}=-1$ 吗?

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乌贼 Posted at 2014-1-14 17:42:08
Last edited by 乌贼 at 2014-1-14 17:50:00回复 18# kuing
有啊,我又看一遍还是$-1$

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kuing Posted at 2014-1-14 17:43:43
回复 19# 乌贼

看清楚题目,人家说的是“存在 $1\leqslant k\leqslant m$,使得 $a_{2k-1}/a_{2k}\ne1$”

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