一道高中组合数学(数数问题)

2025-02-07 21:49:07
推荐回答(6个)
回答1:

该青蛙每跳1次,只有两种方法(逆时针和顺时针)
所以它跳5次,有2^5=32种跳法
但是如果它跳5次以下就到了D,跳法就少了
其实5次以下的情况只有两种,就是ABCD和AFED
这两种情况都是少跳了2次,每种情况少了2^2-1=3种跳法
所以不同的跳法有32-3*2=26种

回答2:

青蛙从A点开始,往相邻两个顶点B和F跳到D点的次数是相同的,可知有13种跳法,此种方法是解决数量较小排列问题的常用方法之一,优点是把抽象变为直观(列树状图)。

回答3:

2*2*2*1=8

回答4:

26种
解析:青蛙从A点开始,往相邻两个顶点B和F跳到D点的次数是相同的,又青蛙第一次往B方向跳A和C或第一次往F跳E和A,以此推
2*13=26

回答5:

若3次跳到D点 跳法有2种 ABCD AFED
4次不可能跳到D点
5次跳到D点 可看作3次跳到D点经过任意一点时有往返 AFABCD(A点往返) ABABCD(B点往返)
ABCBCD(C点往返) ABCDED(D点往返)
在对称的AFED一样有4种跳法

5次不能到D点
5次不可能跳到A点(偶数次能到)
5次跳到B点的跳法可看作一次跳到B点有两次往返 其中又分为重复往返和不重复往返
重复往返 AFAFAB ABCDCD
不重复往返 AFEFAB ABCDCB 4种
5次不可能跳到C点(偶数次能到)
5次不可能跳到E点(偶数次能到 E点和C点是对称点 情况相同)
5次跳到F点有4种(F点和B点是对称点 情况相同)
所以共有18种跳法

回答6:

跳5次有2*5种 跳ABCD后再跳2步有2*2 跳AFED也有2*2 所以2*5-2x2*2=26

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