11类NCERT解决方案-第7章排列和组合-第7章的其他练习

问题1.从DAUGHTER单词的字母中可以形成2个元音和3个辅音中的每个单词，无论有无含义。

解决方案：

In the word “DAUGHTER”, there are 3 vowels, namely, A, U, E, and 5 consonants, namely, D, G, H, T, R.

The number of ways of selecting 2 vowels out of 3 = ³C₂ = $\frac{3!}{2!1!}$ = 3.

The number of ways of selecting 3 consonants out of 5 = ⁵C₃ = $\frac{5!}{3!2!}$ =10.

Therefore, the number of combinations of 2 vowels and 3 consonants is 3 x 10 = 30.

Now, each of these 30 combinations has 5 letters which can be arranged among themselves in 5! ways. 

Therefore, the required number of different words is 30 x 5! = 30 x 120 = 3600

问题2.一次使用“等式”一词的所有字母可以形成多少个有或没有含义的词，从而使元音和辅音同时出现?

解决方案：

There are 8 different letters in the word “EQUATION”, in which there are 5 vowels, namely, A, E, I, O and U and 3 consonants, namely, Q, T and N. Since the vowels and consonants have to occur together, assume all vowels as one object (AEIOU) and all consonants as another (QTN). So, there are 2 objects now.

Then, permutations of these 2 objects taken all at a time = ²P₂ = 2! =2.

Within vowel group, we have 5! Possible permutations and 3! permutations within the consonants group.

Hence, the required number of permutations = 2! x 5! x 3! = 1440.

问题3.必须由9名男孩和4名女孩组成一个由7人组成的委员会。当委员会由以下人员组成时，可以通过几种方式做到：(i)正好是3个女孩? (ii)至少有3个女孩? (iii)最多3个女孩?

解决方案：

(i) We have to select 7 members from 13 (9 boys + 4 girls). 

If there are exactly 3 girls in each combination, then

The number of ways of selecting 3 girls out of 4 = ⁴C₃ = $\frac{4!}{3!1!}$ = 4

The number of ways of selecting 4 boys out of 9 = ⁹C₄ = $\frac{9!}{4!5!}$ = 126

Hence, total number of ways to form committee with exactly 3 girls = 126 x 4 = 504

(ii) Since, the team has to consist of at least 3 girls, the team can consist of 

3 girls and 4 boys, or 4 girls and 3 boys.

Case 1: The team can consist of 3 girls and 4 boys:

So, from(i), we have,

Total number of ways to select 3 girls and 4 boys = 126 x 4 = 504

Case 2: The team can consist of 4 girls and 3 boys:

Total number of ways to select 4 girls and 3 boys = ⁴C₄ x ⁹C₃ = 1 x 84 = 84

Therefore, total number of ways to form a committee with at least 3 girls = 504 + 84 = 588 

(iii) Since, the team has to consist of at most 3 girls, the team can consist of

3 girls and 4 boys, or 2 girls and 5 boys, or 1 girl and 6 boys, or 7 boys

Case 1: The team can consist of 3 girls and 4 boys

Total number of ways to select 3 girls and 4 boys = 126 x 4 = 504

Case 2: The team can consist of 2 girls and 5 boys

Total number of ways to select 2 girls and 5 boys = ⁴C₂ x ⁹C₅ = $\frac{4!}{2!2!} \times \frac{9!}{5!4!}$ = 126 x 6 = 756

Case 3: The team can consist of 1 girl and 6 boys

Total number of ways to select 1 girl and 6 boys = ⁴C₁ x ⁹C₆ = $\frac{4!}{1!3!} \times \frac{9!}{6!3!}$ = 84 x 4 = 336

Case 4: The team can consist of 7 boys

The number of ways of selecting 7 boys out of 9 = ⁹C₇ = $\frac{9!}{7!2!}$ = 36

Therefore, total number of ways to form a committee with at most 3 girls = 504 + 756 + 336 + 36 = 1632

问题4.如果在词典中列出了单词EXAMINATION的所有字母的不同排列，那么在列表中以E开头的第一个单词之前有多少个单词?

解决方案：

In a dictionary, words are in ascending order. Hence, for this scenario, all the words starting with A will come before first word starting with E. 

To get the number of words starting with A, we fix the letter A at the extreme left position and then rearrange the remaining 10 letters taken all at a time.  These 10 letters includes I and N 2 times.

Hence, the number of words starting with A = $\frac{10!}{2!2!}$ = 907200. 

Hence, 907200 words are there in the list before the first word starting with E.

问题5.由数字0、1、3、5、7和9可以形成多少个6位数字，这些数字可以被10整除，并且没有重复的数字?

解决方案：

We know that, a number is divisible by 10 if it has 0 at its unit place. Hence, 0 will be fixed at last place in a number. 

The remaining 5 places can be filled by any digit 1, 3, 5, 7 or 9. These 5 digits can be arranged at 5 places in ⁵P₅ = 5! ways.

Hence, 6-digit numbers that can be formed which are divisible by 10 = 5! = 120.

问题6.英文字母有5个元音和21个辅音。字母表可以形成多少个带有两个不同元音和两个不同辅音的单词?

解决方案：

We have to select 2 vowels from 5 and 2 consonants from 21.

The number of ways to select 2 vowels from 5 = ⁵C₂ = $\frac{5!}{2!3!}$ = 10

The number of ways to select 2 consonants from 21 = ²¹C₂ = $\frac{21!}{2!19!}$ = 210

Permutations of these 4 alphabets taken all at a time = ⁴P₄ = 4! = 24.

Hence, number of words can be formed with 2 different vowels and 2 different consonants = 10 x 210 x 24 = 50400

问题7.在考试中，试卷由12个问题组成，分为两个部分，即第一部分和第二部分，分别包含5和7个问题。要求一个学生总共尝试8个问题，并从每个部分中至少选择3个。学生可以通过几种方式选择问题?

解决方案：

Since, it is required to attempt at least 3 questions from each part, questions selection can be  

Case 1: (3, 5), or (b)(4, 4), or (c) (5, 3) ……(Part I questions, Part 2 questions)

Part I consists of 5 questions and Part II consists of 7 questions.

So, Number of ways to select 3 questions from Part I = ⁵C₃ = $\frac{5!}{3!2!}$ = 10

Number of ways to select 5 questions from Part II = ⁷C₅ = $\frac{7!}{5!2!}$ = 21  

Hence, total number of ways for this type of question selection = 10 x 21 = 210

Case 2: Number of ways to select 4 questions from Part I = ⁵C₄ = $\frac{5!}{4!1!}$ = 5

Number of ways to select 4 questions from Part II = ⁷C₄ = $\frac{7!}{4!3!}$ = 35

Hence, total number of ways for this type of question selection = 5 x 35 = 175

Case 3: Number of ways to select 5 questions from Part I = ⁵C₅ = 1

Number of ways to select 3 questions from Part II = ⁷C₃ = $\frac{7!}{3!4!}$ = 35

Hence, total number of ways for this type of question selection = 1 x 35 = 35

Therefore, a student can select the questions in 210 + 175 + 35 = 420 ways.

问题8.如果选择的5张卡中的每一个都有一个国王，则从52张卡中确定5张卡组合的数量。

解决方案：

We have to select 5 cards from 52 cards. If there is exactly one king in each combination, then

Case 1: We have to select 1 king card from 4 king cards

Number of ways to select king card = ⁴C₁ = $\frac{4!}{1!3!}$ = 4

Case 2: We have to select 5 – 1 = 4 cards from remaining 52 – 4 = 48 cards

Number of ways to select remaining 4 cards = ⁴⁸C₄ = $\frac{48!}{4!44!}$ = 194580

And, hence required total number of 5 card combinations = 4 x 194580 = 778320.

问题9.必须连续坐5名男性和4名女性，以使女性占据均匀的位置。可能有多少种这样的安排?

解决方案：

There are 5 men and 4 women. Women occupy the even places. Hence, the possible seating arrangement of men and women will be MWMWMWMWM where M denotes Man and W denotes Woman.

Now, 4 women can occupy 4 even places in ⁴P₄ = 4! ways.

And, 5 men can occupy 5 odd places in ⁵P₅ = 5! ways.

Hence, Total number of such possible arrangements: 

5! x 4! = 5 x 4 x 3 x 2 x 1 x 4 x 3 x 2 x 1 = 2880.

问题10.从25名学生中选出10人参加郊游聚会。有3个学生决定他们所有人都将加入或不参加。可以选择多少种游览方式?

解决方案：

According to those 3 students who decide to either join all of them or join none of them, there are 2 cases for excursion party member selection:

Case 1: Do not select those 3 students. 

Hence, select 10 students from remaining 25 – 3 = 22 students.

The number of ways to do this = ²²C₁₀ = 646646

Case 2: Select those 3 students. 

This means, we are short of 7 students and these can be selected from remaining 25 – 3 = 22 students.

The number of ways to do this = ²²C₇ x ³C₃ = 170544

Hence, required number of ways to choose excursion party:

646646 + 170544 = 817190

问题11.可以用多少种方式排列“暗杀”一词的字母，以使所有的S在一起?

解决方案：

Word “ASSASSINATION” has 4 S. We have to place all S’s together. Hence, we will assume group of 4 S as a single object. This single object together with 9 remaining letters (objects) will be counted as 10 objects to be arranged. These include 3 A’s, 2 I’s and N’s and 1 T and O. 

So, required number of ways = $\frac{10!}{3!2!}$ = 151200