第11类RD Sharma解决方案–第17章组合-练习17.3

问题1：用5个元音和17个辅音可以形成多少个不同的单词，每个单词包含2个元音和3个辅音。

解决方案：

Given, word contains 2 vowels and 3 consonants.

So, we need to choose 2 vowels out of 5 vowels and 3 consonants out of 17 consonants.

This can be done in $^5C_3.^{17}C_3$ ways.

Also, we need number of different words, we can arrange 5 letter word in 5! ways.

Therefore, Total number of ways = $(^5C_3.^{17}C_3)5!$

⇒ 6800 × 120 = 816000

问题2：有10个人命名为P ₁ ，P ₂ ，P ₃ ….P ₁₀人中有10个人，应将5个人排成一排，这样在每次安排中都必须出现_{P 1，而P 4}和P ₅不会发生。查找此类可能安排的数量。

解决方案：

Given, we need to arrange 5 persons out of 10 persons such that in each arrangements P₁ must occur whereas P₄ and P₅ do not occur.

Here, we Should choose P₁ every time, so now, we choose 4 persons out of 9 persons.

From that 9 persons, we don’t need to choose P₄ and P₅, so This can be done in $^7C_4$ ways.

Therefore, number of selections = $^7C_4$

And, 5 persons can be arranged in 5! ways. so,

Total number of ways = $^7C_4.5!$

⇒ 4200.

Therefore, number of such possible arrangements is 4200.

问题3：假设没有字母重复，那么“ MONDAY”一词的字母可以形成多少个有意义或没有意义的单词。如果

(i)一次使用4个字母。

解决方案：

Given, six letter word “MONDAY”.

4 letters are used at a time out of 6 letters. this can be done in $^6C_4$ ways.

These four letters can be arranged in 4! ways.

Therefore, total number of ways = $^6C_4.4!$

⇒ 360.

(ii)一次使用所有字母。

解决方案：

Given, six letter word “MONDAY”.

6 letters are used at a time out of 6 letters. this can be done in $^6C_6$ ways.

These 6 letters can be arranged in 6! ways.

Therefore, Total number of ways = $^6C_6.6!$

⇒ 1 × 720 = 720

(iii)所有字母均已使用，但首字母为元音。

解决方案：

Given, six letter word “MONDAY”.

All letters are used at a time but first letter is Vowel,

number of vowels in word “MONDAY” is 2. So first we choose one vowel from these two in $^2C_1$ ways.

The remaining 5 letters out of 5 letters. This can be done in $^5C_5$ ways.

Number of arrangements can be done for these 5 letters are 5!.

Therefore, total number of ways = $^2C_1.^5C_5.5!$

⇒ 2\times1\times120=240.

问题4：求n个不同事物在r中合在一起的排列数量，其中3个特定事物必须一起出现。

解决方案：

Given, Number of permutations of n distinct things taking r together and 3 particular things are already selected.

So, now number of ways of choosing (r – 3) things from remaining (n – 3) things is,

 This can be done in $^{n-3}C_{r-3}$ ways.

We need to find the number of permutations, there are total (r – 2) things considering 3 particular things as single thing.

These can be arranged in (r – 2)! ways.

Internally, 3 particular things can be arranged in 3! ways.

Therefore, Total number of permutations = $^{n-3}C_{r-3}\times(r-2)!\times3!$

问题5：从INVOLUTE单词的字母中可以形成每个3个元音和2个辅音多少个单词?

解决方案：

Given word is “INVOLUTE”.

Number of vowels and consonants in the word are 4 and 4 respectively.

Number of ways in choosing 3 vowels out of 4 and 2 consonants out of 4 is $^4C_3\times^4C_2$

These five letters can be arranged in 5! ways.

Therefore, Total number of words formed = $^4C_3\times^4C_2\times5!$

⇒ 4 × 6 × 120 = 2880

问题6：找出一次r使得n个不同事物同时发生的n个不同事物的排列数量?

解决方案：

Given, number of permutations of n different things taken r at a time and two specified things occur together.

So, we now choose (r – 2) things from the remaining (n – 2) things.

This can be done in $^{n-2}C_{r-2}$ ways.

We need to find the number of permutations, there are total (r – 1) things considering 2 specific things as single thing.

These can be arranged in (r – 1)! ways.

Internally, 2 particular things can be arranged in 2! ways.

Therefore, Total number of permutations = $^{n-2}C_{r-2}\times(r-1)!\times2!$

⇒ $2(r - 1)\times ^{n-2}C_{r-2}\times(r-2)!$

⇒ $2(r - 1)\times ^{n-2}P_{r-2}$

问题7：找出以下几种方式：(a)选择(b)排列，可以从“比例”一词的字母中选出四个字母。

解决方案：

Given word is ‘PROPORTION’

there are 10 letters in this word and mainly ‘OOO’, ‘PP’, ‘RR’, ‘I’, ‘T’, ‘N’. 

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct 

here we find that only one 3 alike letter in the word (‘OOO’).

So, choosing 1 distinct letter from remaining 5 distinct letters is $^5C_1$

⇒ 5 ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters , number of ways in choosing 1 letters out of 3 is $^3C_1$ 

And number of ways of choosing 2 distinct letters from remaining 5 letters is $^5C_2$

⇒ $^3C_1\times ^5C_2=30$  ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is $^3C_2$

⇒ 3 ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is $^6C_4$

⇒ 15 ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 5+30+3+15=53 ways.

(b) Here we need to arrange 4 letters out of 10 letters, here cases will be same as selection, but we arrange the given letters in every case.

case-1: Number of ways of selecting 3 alike and 1 distinct letters is 5.

Number of ways in arranging is similar to arranging the n people in n places where r places are same = $\frac{n!}{r!}$

similarly, arrangement of 4 letters where 3 are alike is $\frac{4!}{3!}$

total number of ways are $5\times \frac{4!}{3!} =20$

case-2: Number of ways of selecting 2 alike letters and 2 distinct letters is 30 ways.

number ways of arranging them is $\frac{4!}{2!}$

Total number of ways are $30\times \frac{4!}{2!} =360$

case-3: Number of ways of selecting 2 letters alike and 2 letters alike is 3 ways.

Number of ways in arranging them is $\frac{4!}{2!\times2!}$

Total number of ways are $3\times \frac{4!}{2!\times2!} =18$

Case-4: Number of ways of selecting 4 distinct letters is 15 ways.

number of ways of arranging them is 4!

Total number of ways are 15 × 4! = 360

considering all cases, Total number of ways are sum of all number of ways of all cases

⇒ 20+360+18+360 = 758 ways.

问题8：从“ MORADABAD”单词的字母中一次取4个字母可以形成多少个单词?

解决方案：

Given word is ‘MORADABAD’

There are 10 letters in this word and mainly ‘AAA’, ‘DD’, ‘M’, ‘R’, ‘B’, ‘O’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 3 alike letters and 1 distinct letter
2. 2 alike letters and 2 distinct letters
3. 2 alike letters of I kind and 2 alike letters of other kind.
4. all distinct letters.

lets consider,

case-1: 3 alike and 1 distinct

here we find that only one 3 alike letter in the word (‘AAA’).

So, choosing 1 distinct letter from remaining 5 distinct letters is $^5C_1$

Number of ways of arranging them is $\frac{4!}{3!}$

⇒ $5\times\frac{4!}{3!}=20$ ways.

case-2: 2 alike letters and 2 distinct letters.

Here, we find that 2 chars have 2 alike letters , number of ways in choosing 1 letters out of 2 is $^2C_1$

And number of ways of choosing 2 distinct letters from remaining 5 letters is $^5C_2$

Number of ways of arranging them is $\frac{4!}{2!}$

⇒ $^2C_1\times ^5C_2\times\frac{4!}{2!}=240$ ways.

case-3 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 2 chars have 2 alike letters, number of ways in choosing 2 letters out of 2 is $^2C_2$

Number of ways of arranging them is $\frac{4!}{2!\times2!}$

⇒ $^2C_2\times\frac{4!}{2!\times2!}=6$ ways

Case-4: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is $^6C_4$

Number of ways of arranging them is 4!

⇒ $^6C_4\times4!= 360$ ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 20+240+6+360 = 626 ways.

问题9：一位商人为21位客人举办晚宴。他有2个圆桌会议，每个可容纳15和6人。他可以用几种方式安排客人?

解决方案：

Given, A business man hosts dinners to 21 guests, where 2 round tables accommodate 16 and 6 persons.

So, choosing 15 guests out of 21 to accommodate in one table in $^{21}C_{15}$ ways.

Those 15 guests can be arranged in themselves in (15 – 1)! ways. because it is a round table. 

So, we need to keep a guest fixed and arrange remaining 14 guests = (15 – 1)! = 14!.

After accommodating 15 guests in one table, accommodating remaining 6 guests out of 6 in another table in $^6C_6$ ways.

Those 6 guests can be arranged themselves in (6 – 1)! = 5! ways.

Total number of ways = $^{21}C_{15}\times^6C_6\times14!\times5!=^{21}C_{15}\times14!\times5!$

问题10：找出从“考试”一词中提取的4个字母的组合和排列的数量。

解决方案：

Given word is ‘EXAMINATION’

There are 10 letters in this word and mainly ‘AA’, ‘NN’, ‘II’, ‘E’, ‘X’, ‘O’,’M’,’T’.

(a) Here we need to select 4 letters out of 10 letters. but we need to consider some cases.

1. 2 alike letters and 2 distinct letters
2. 2 alike letters of I kind and 2 alike letters of other kind.
3. all distinct letters.

lets consider,

case-1: 2 alike letters and 2 distinct letters.

Here, we find that 3 chars have 2 alike letters, number of ways in choosing 1 letters out of 3 is $^3C_1$

And number of ways of choosing 2 distinct letters from remaining 7 letters is $^7C_2$

Number of ways of arranging them is $\frac{4!}{2!}$

⇒ $^3C_1\times ^7C_2\times\frac{4!}{2!}=756$ ways.

case-2 : 2 alike letters of I kind and 2 alike letters of other kind.

Here, we find 3 chars have 2 alike letters, number of ways in choosing 2 letters out of 3 is $^3C_2$

Number of ways of arranging them is $\frac{4!}{2!\times2!}$

⇒ $^3C_2\times\frac{4!}{2!\times2!}=18$ ways

Case-3: all distinct letters.

Here, we have 6 different letters, number of ways in choosing 4 letters out of 8 is $^8C_4$

Number of ways of arranging them is 4!

⇒ $^8C_4\times4!= 1680$ ways.

Therefore, Total number of ways is sum of number of ways in all cases.

⇒ 756+18+1680 = 2454 ways.

问题11：在一张长桌的两边安排了16人的茶会，每边有8张椅子。四人希望坐在一个特定的一侧，而另两个人则坐在另一侧。他们可以以几种方式就座?

解决方案：

Given, A tea party is arranged 16 persons along two sides of long table with 8 chairs on each side.

Let two sides be side A, side B.

Also, 4 persons wish to sit on side A, and 2 persons on side B.

Remaining seats in side A and side B are 4, 6 respectively.

Now, let we choose 4 persons outs of remaining 10 on side A. 

This can be done in $^{10}C_4$ ways.

Also, choosing 6 persons out of remaining 6 on side B.

This can be done in $^6C_6$ ways.

Now, 8 persons on each side can be arranged in 8! ways.

Therefore, total number of ways = $^{10}C_4\times^6C_6\times8!\times8!$

⇒ $^{10}C_4\times(8!)^2$