节: 因数和组合 | 7.11 因数和组合

章节大纲

Kelly and Kyle are playing a card game, and Kyle is wondering why there never seem to be repeated hands. He figures that since there are only 52 cards in the deck, and each hand has five cards, there really should be more duplicate hands. After all, 5 is nearly 1/10 of 52.
::Kelly和Kyle在玩牌游戏,Kyle想知道为什么从未出现过重复手。他估计甲板上只有52张牌,每张牌有5张牌,真的应该有更多的重复手。毕竟,5张几乎是52张的1/10。

Kelly tells him that she thinks there are thousands of possible combinations, and that she would be really surprised to see the same 5 card hand twice or more in a given game.
::Kelly告诉他,她认为有成千上万的组合, 她会非常惊讶地看到同样的5张纸牌手两次或两次以上的游戏。

Who is correct? Why?
::谁是正确的为什么

Factorials and Combinations
::因数和组合

Recall that a factorial of a positive integer n is the product of n , and all of the positive integers less than n . We write this as n ! = n ( n - 1)( n - 2) .. (3) (2) (1).
::回顾正整数的乘数 n 是 n 的产物, 所有正整数小于 n。我们将此写为 n! = n( n) - 1 - 2 . (3) (2) (1)。

In order to develop the binomial theorem, we need to look at a related idea: combinations. If you have studied probability , you may be familiar with combinations and permutations. A is the number of ways you can choose r objects from a group of n objects, if the order of choosing does not matter. The following examples will help clarify the idea of a combination.
::为了开发二进制定理, 我们需要查看一个相关的想法: 组合。如果您研究过概率, 您可能熟悉组合和变换。 A 是您可以从 n 对象组中选择 r 对象的数, 如果选择的顺序无关紧要。以下示例将有助于澄清组合的概念。

Examples
::实例

Example 1
::例1

Earlier, you were asked a question about the disagreement between Kelly and Kyle.
::之前有人问过你 Kelly和Kyle之间的分歧

Kelly thinks there are thousands of possible 5-card combinations in a deck of cards, Kyle thinks there should not be all that many.
::Kelly认为牌牌甲板上可能有数千张5卡组合凯尔认为不应该有那么多

This is a classic combinations problem in the form "52, choose 5":
::这是一个典型的组合问题, 以“52, 选择 5” 的形式出现:

Using the formula: $\begin{align*}_{n}C_{r} = \binom{n} {r} = \frac{n!} {r!(n - r)!}\end{align*}$
::使用公式: nCr=(nr)=n!r!!

We get: $\begin{align*}_{52}C_{5} = \binom{52} {5} = \frac{52!} {5!(52 - 5)!}\end{align*}$
::我们得到:52C5=(525)=52!5!(52-5)!

Simplify to: $\begin{align*}\frac{52!} {5!(47)!}\end{align*}$
::简化为: 52! 5! (47)!

Simplify more: $\begin{align*}\frac{52 \cdot 51 \cdot 50 \cdot 49 \cdot 48} {5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}\end{align*}$
::简化更多: 525150494854321

And once more: $\begin{align*}\frac{311,875,200}{120}\end{align*}$
::再来一次:311 875 200 120

Gives: $\begin{align*}2,598,960\end{align*}$
::给予:2,598,960

Looks like Kelly undershot by quite a bit too!
::看来凯莉的枪法也太低了!

Example 2
::例2

In a class of 20 students, 3 students are going to be chosen to form a committee to plan a fieldtrip. How many possible committees are there?
::在20名学生的班级中,将挑选3名学生组成一个计划野外考察的委员会。可能有多少委员会?

To answer this question, we need to figure out how many ways we can choose groups of 3 students from the 20 on the class. The order of choosing does not matter. That is, if I choose Amy, Juan, and Nina, it is the same as choosing Juan, then Amy, then Nina, or any other ordering of the three students.
::为了回答这个问题,我们需要弄清楚我们有多少方法可以选择班上20个学生中的3个。选择顺序无关紧要。这就是说,如果我选择了艾米、胡安和尼娜,这与选择胡安、然后是艾米、然后是尼娜或三个学生的任何其他顺序是一样的。

In general, we can find the number of combinations of r objects chosen from n objects by the following:
::一般而言,我们可以找到从 n 对象中选择的r 物体的组合数,具体如下:

	$\begin{align}_{n}C_{r} = \binom{n} {r} = \frac{n!} {r!(n - r)!}\end{align}$

(Note that there are two different symbols for combinations: $\begin{align*}_{n}C_{r}\end{align*}$ and $\begin{align*}\binom{n} {r}\end{align*}$ You can use either one, though $\begin{align*}_{n}C_{r}\end{align*}$ is what is used on the TI-83/84.
:注意组合有两个不同的符号: nCr 和 (nr) 您可以使用其中之一, 尽管 NCr 是 TI-83/84 上所用的符号。)

Therefore the number of combinations of 3 people from 20 people in the class is
::因此,班级中20人中的3人组合数是:

	$\begin{align}\frac{20!} {3!17!} = 1140\end{align}$

Example 3
::例3

How many different groups of 3 cards can be chosen from 20 different cards, assuming order does not matter? Use a graphing calculator.
::从20张不同的卡片中可以选取多少组不同的3张卡片,假设顺序不重要?使用图形计算器。

To find $\begin{align*}\binom{20} {3} \end{align*}$ :
::要查找( 203) :

Press: 20 and then move right to the PRB menu.
::按下: 20 然后右移到 PRB 菜单。

Press 3. This takes you back to the main screen. You should see 20 $\begin{align*}_{n}C_{r}\end{align*}$ .
::按3, 这带您返回主屏幕。您应该看到 20 nCr 。

Now press 3 >,TI font_ENTER>.
::按 3 >,TI 字体_ENTER>。

You should see the answer, 1140.
::你应该看看答案,1140

Example 4
::例4

Calculate by hand: How many different 4-person teams can be made from 7 people?
::由7人组成多少个不同的四人小组?

Smaller numbers, such as these, are not too difficult to calculate by hand.
::象这些小数字这样的小数字并非很难用人工计算。

$\begin{align*}\binom{7} {4} = \frac{7!} {4!3!} = \frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1} {4 \cdot 3 \cdot 2 \cdot 1 \cdot 3 \cdot 2 \cdot 1} = \frac{7 \cdot 6 \cdot 5} {3 \cdot 2 \cdot 1} = 7 \cdot 5 = 35\end{align*}$ .

Canceling factors in the numerator and denominator simplifies the calculation.
::分子和分母中的取消系数简化计算。

Example 5
::例5

Simplify: $\begin{align*}\frac{(m + 3)!}{m + 1}!\end{align*}$
::简化m+3)!m+1!

First we need to expand the numerator and denominator:
::首先,我们需要扩展分子和分母:

$\begin{align*}\frac{(m + 3)(m + 2)(m + 1)(m)(m - 1)...(1)}{(m + 1)(m)(m - 1)...(1)}\end{align*}$
:m+3)(m+2)(m+1)(m)(m)(m-1).(1)(m+1)(m)(m)(m-1).(1)。

$\begin{align*}(m + 3)(m + 2)\end{align*}$ ..... Cancel common factors
:m+3(m+3)(m+2))..

$\begin{align*} m^2 + 5m +6\end{align*}$ ..... Simplify
::m2+5m+6.... 简化

Example 6
::例6

Demonstrate that: $\begin{align*}_{2(4)}C_2 = 2(_4C_2) + 4^2\end{align*}$ .
::证明:2(4)C2=2(4C2)+42。

Evaluate both sides:
::评估双方:

$\begin{align*}_8C_2 = 28\end{align*}$
::8C2=28

$\begin{align*}_4C_2 = 6\end{align*}$
::4C2=6

$\begin{align*} 4^2 = 16\end{align*}$

Check: $\begin{align*}28 = 2(6) + 16\end{align*}$
::检查: 28=2(6)+16

$\begin{align*}28 = 12 + 16\end{align*}$

$\begin{align*}28 = 28\end{align*}$ - the equality holds.
::28=28 - 人人平等。

Example 7
::例7

In how many ways can I pick 6 jelly beans from a container containing 10 jelly beans?
::我能从装有10个果豆的集装箱里采摘多少种豆子?

We calculate using the formula: $\begin{align*}_{10}C_6 = \frac{10!}{6!(10 - 6)!}\end{align*}$
::我们用公式计算: 10C6=10! 6!! (10- 6)!

$\begin{align*}10! = (1)(2)(3)(4)(5)(6)(7)(8)(9)(10)\end{align*}$

$\begin{align*}10! = 3628800\end{align*}$

$\begin{align*}6! = (1)(2)(3)(4)(5)(6)\end{align*}$

$\begin{align*}6! = 720\end{align*}$

$\begin{align*}(10 - 6)! = (1)(2)(3)(4)\end{align*}$

$\begin{align*}(10-6)! = 24\end{align*}$

$\begin{align*}\therefore _{10}C_6 = \frac{3628800}{720 \cdot 24} \to 210\end{align*}$
::10C6=362880072024210

Example 8
::例8

At the carnival, you decide to play a game of chance. You buy 15 tickets for the game. You have a 75% chance of winning each time you play the game. What is the probability that you will win exactly 8 of the 15 games?
::在嘉年华会上,你决定玩一个机会游戏。你买15张比赛票。每次玩游戏,你都有75%的胜算。在15场比赛中,你赢得8场的几率是多少?

The chance is about 4%.
::机会大约是4%。

For each of the 15 games, there is 75% chance of winning, and a 25% chance of losing.
::在15场比赛中,每场比赛都有75%的胜选机会,25%的败赛机会。

The probability of exactly 8 wins is $\begin{align*}\mathit \ \binom{15}{8}(.75)^{8}(.25)^{7}\approx 0.039 \to 3.9\%\end{align*}$
::8胜的概率是 (158) 758 (.25)7 0.0393.9%

Note that this is only the probability of winning exactly 8 games, no more, no less.
::请注意,这只是8场比赛完全胜出的可能性,不胜枚举,不减不减。

Example 9
::例9

Explain why the following equality holds: $\begin{align*}_{(n+1)}C_r = _nC_r + _n C _{r-1}\end{align*}$ .
::解释为何存在以下平等n+1)Cr=nCr+nCr-1。

Proof of equality:
::平等的证据:

First, remove one item from the set.
::首先,从集中删除一个项目。

Then, either the (r) item we want to come out of the remaining items, or we choose (r - 1) item from the remaining items and we include the one item we removed.
::然后,要么(r)项我们想从其余项中拿出来,要么我们从项中选择(r-1)项,我们列入一个我们删除的项目。

Review
::回顾

Simplify and evaluate the factorials.
::简化和评估阶乘。

$\begin{align*}\frac{(b - 2)!}{(b - 5)!}\end{align*}$
:b -2)! (b -5)! (b -5)!

$\begin{align*}\frac{(a + 2)!}{(a + 1)!}\end{align*}$
:a+2)!(a+1)!
$\begin{align*}\frac{7!}{3! 3!}\end{align*}$
$\begin{align*}\frac{9!}{8!}\end{align*}$
$\begin{align*} 4! + 3!\end{align*}$
$\begin{align*}\frac {6!}{5!}\end{align*}$

Show that the equality holds by evaluating both sides of the equation:
::通过评价等式的两面来显示平等是站得住脚的:

$\begin{align*}_4C_2 = _4C_{4-2}\end{align*}$
::4C2=4C4-2

$\begin{align*}_{(7+1)}C_4=_7C_4 + _7C_{(4-1)}\end{align*}$
:7+1)C4=7C4+7C(4-1)

Explain conceptually why the following equality holds:
::从概念上解释以下平等为何存在:

$\begin{align*}_{2n}C_2 = 2(_nC_2) + n^2\end{align*}$
::2nC2=2(nC2)+n2

Solve.
::解决。

$\begin{align*} _{8}C_{5}\end{align*}$
::8C58C5

$\begin{align*} _{6}C_{3}\end{align*}$
::6C3 6C3

In a class of 200 students, 25 will be chosen randomly to participate in a research study. How many possible groups of 25 students can be chosen? (Hint: use a calculator!)
::在200名学生的班级中,将随机挑选25人参加一项研究。可以选择多少组25名学生? (提示:使用计算器! )

A die is rolled 10 times. What is the probability of rolling exactly four 4’s? (Hint: the probability of rolling a 4 is 1/6. The probability of not rolling a 4 is 5/6.)
::死亡是滚动的10倍。滚动的概率是多少? (时间:滚动的概率是1/6。不滚动的概率是5/6。 )

The local TV station forecasts a 30% chance of rain every day for the next week. What is the probability that it will rain on exactly 6 out of the next 7 days?
::当地电视台预测下星期每天有30%的雨量。未来7天中,有6天的雨量会有多大的概率?

Consider the following situation: a basketball player is going to attempt to make 20 free throws. She is assuming that she has an 80% chance of making each shot. What is the probability that she will make exactly 19 out of 20 shots?
::考虑一下以下情形:篮球运动员将尝试免费投球20次。她假设她有80%的投篮机会。在20次投篮中,她能射出19次的概率是多少?

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

章节大纲

Factorials and Combinations ::因数和组合

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Example 6 ::例6

Example 7 ::例7

Example 8 ::例8

Example 9 ::例9

Review ::回顾

Review (Answers) ::回顾(答复)

Resources ::资源