有两个独特的盒子,10个相同的红色球,10个相同的黄色球,10个相同的蓝色球。有多少种方法可以将30个球分类为两个盒子,以便每个盒子都有15个?据/h3>
记住这两个盒子是不同的,让据span class="katex">
R.据/span>那据/span>y据/span>和据span class="katex">
B.据/span>分别为第一个方框中红色、黄色和蓝色球的数字。然后我们首先需要得到满足的案例数据span class="katex">
R.据/span>+据/span>y据/span>+据/span>B.据/span>=据/span>1据/span>5.据/span>那据/span>然后减去这些情况的个数据span class="katex">
R.据/span>>据/span>1据/span>0.据/span>那据/span>y据/span>>据/span>1据/span>0.据/span>或据span class="katex">
B.据/span>>据/span>1据/span>0.据/span>.据/span>
使用星星和酒吧,案件令人满意的数量据span class="katex">
R.据/span>+据/span>y据/span>+据/span>B.据/span>=据/span>1据/span>5.据/span>是据span class="katex">
(据/span>2据/span>1据/span>7.据/span>)据/span>=据/span>1据/span>3.据/span>6.据/span>.据/span>(据/span>1据/span>)据/span>
下面给出了这种情况的数量据span class="katex">
1据/span>0.据/span>据据/span>R.据/span>≤据/span>1据/span>5.据/span>:据/span>
- 如果据span class="katex">
R.据/span>=据/span>1据/span>1据/span>那据/span>然后据span class="katex">
y据/span>+据/span>B.据/span>=据/span>4.据/span>那据/span>暗示有5个这样的例子。据/li>
- 如果据span class="katex">
R.据/span>=据/span>1据/span>2据/span>那据/span>然后据span class="katex">
y据/span>+据/span>B.据/span>=据/span>3.据/span>那据/span>暗示有4个这样的情况。据/li>
- 如果据span class="katex">
R.据/span>=据/span>1据/span>3.据/span>那据/span>然后据span class="katex">
y据/span>+据/span>B.据/span>=据/span>2据/span>那据/span>暗示有3个这样的情况。据/li>
- 如果据span class="katex">
R.据/span>=据/span>1据/span>4.据/span>那据/span>然后据span class="katex">
y据/span>+据/span>B.据/span>=据/span>1据/span>那据/span>这意味着有两种情况。据/li>
- 如果据span class="katex">
R.据/span>=据/span>1据/span>5.据/span>那据/span>然后据span class="katex">
y据/span>+据/span>B.据/span>=据/span>0.据/span>那据/span>暗示有1个这样的例子。据/li>
因此,其中的数字情况据span class="katex">
1据/span>0.据/span>据据/span>R.据/span>≤据/span>1据/span>5.据/span>是据span class="katex">
5.据/span>+据/span>4.据/span>+据/span>3.据/span>+据/span>2据/span>+据/span>1据/span>=据/span>1据/span>5.据/span>.据/span>因为完全相同的逻辑适用于据span class="katex">
1据/span>0.据/span>据据/span>y据/span>≤据/span>1据/span>5.据/span>或据span class="katex">
1据/span>0.据/span>据据/span>B.据/span>≤据/span>1据/span>5.据/span>那据/span>减去的总数据span class="katex">
(据/span>1据/span>)据/span>是据span class="katex">
3.据/span>×据/span>1据/span>5.据/span>=据/span>4.据/span>5.据/span>.据/span>(据/span>2据/span>)据/span>
因此,服用据span class="katex">
(据/span>1据/span>)据/span>−据/span>(据/span>2据/span>)据/span>给我们的答案据span class="katex">
1据/span>3.据/span>6.据/span>−据/span>4.据/span>5.据/span>=据/span>9.据/span>1据/span>.据/span>
□据/span>
PizzaHot制作7种披萨,其中3种每周7天特价。根据他们的政策,任何两种在同一天打折的披萨在日历周剩下的时间里都不能在同一天打折。让据span class="katex">
X据/span>是一个日历周内所有可能的销售策略的数量。剩下的是什么据span class="katex">
X据/span>划分为1000?据/h3>
让据span class="katex">
一种据/span>那据/span>B.据/span>那据/span>C据/span>那据/span>D.据/span>那据/span>E.据/span>那据/span>F据/span>那据/span>G据/span>是7种披萨。那么这7个都不是每周待售4天或更长时间的,因为其他6种中的每一个都会在前3天的同一天出售。事实上,既然存在据span class="katex">
3.据/span>×据/span>7.据/span>=据/span>2据/span>1据/span>每周有7种披萨促销,每周三次。据/p>
现在,不失一般性,选择3天放多少种方法据span class="katex">
一种据/span>销售是据span class="katex">
(据/span>3.据/span>7.据/span>)据/span>.据/span>那么剩下的6种商品在这3天内的销售方式是据span class="katex">
(据/span>2据/span>6.据/span>)据/span>×据/span>(据/span>2据/span>4.据/span>)据/span>×据/span>(据/span>2据/span>2据/span>)据/span>.据/span>下表是该操作的一个示例,其中据span class="katex">
一种据/span>本周所有3天都在销售,而其他6种的每一个仅在一次销售中:据/p>
披萨据/span>
既然我们结束了据span class="katex">
一种据/span>那据/span>现在我们放一个据span class="katex">
B.据/span>在剩下的2天中的每一个中,做的方式数量是据span class="katex">
(据/span>2据/span>4.据/span>)据/span>.据/span>然后,不包括据span class="katex">
C据/span>哪个已经在销售据span class="katex">
B.据/span>第一天,我们把据span class="katex">
D.据/span>那据/span>E.据/span>那据/span>F据/span>那据/span>G据/span>在2天的地方据span class="katex">
B.据/span>只是放了。但是,自组合以来据span class="katex">
(据/span>D.据/span>那据/span>E.据/span>)据/span>和据span class="katex">
(据/span>F据/span>那据/span>G据/span>)据/span>已经被用来处理据span class="katex">
一种据/span>那据/span>摆放方式的数量据span class="katex">
D.据/span>那据/span>E.据/span>那据/span>F据/span>那据/span>G据/span>在两列一起据span class="katex">
B.据/span>是据span class="katex">
(据/span>2据/span>4.据/span>)据/span>−据/span>2据/span>.据/span>
最后,我们放一个据span class="katex">
C据/span>在剩下的两列中然后填满这两列,有2种方法。因此,在一个日历周内所有可能的销售策略的数量是据span class="katex-display">
(据/span>3.据/span>7.据/span>)据/span>×据/span>(据/span>2据/span>6.据/span>)据/span>×据/span>(据/span>2据/span>4.据/span>)据/span>×据/span>(据/span>2据/span>4.据/span>)据/span>×据/span>(据/span>(据/span>2据/span>4.据/span>)据/span>−据/span>2据/span>)据/span>×据/span>2据/span>=据/span>1据/span>5.据/span>1据/span>2据/span>0.据/span>0.据/span>.据/span>
因此,剩下的据span class="katex">
1据/span>5.据/span>1据/span>2据/span>0.据/span>0.据/span>划分为1000是200。据span class="katex">
□据/span>