# Class 11 RD Sharma Solutions – Chapter 17 Combinations- Exercise 17.3

**Question 1: How many different words, each containing 2 vowels and 3 consonants can be formed with 5 vowels and 17 consonants.**

**Solution:**

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 ways.

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

Therefore, Total number of ways =

â‡’ 6800 Ă— 120 = 816000

**Question 2: There are 10 persons named P**_{1},P_{2},P_{3}….P_{10} out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P_{1} must occur whereas P_{4} and P_{5} do not** occur. Find the number of such possible arrangements.**

_{1},P

_{2},P

_{3}….P

_{10}out of 10 persons, 5 persons are to be arranged in a line such that in each arrangement P

_{1}must occur whereas P

_{4}and P

_{5}

**Solution: **

Given, we need to arrange 5 persons out of 10 persons such that in each arrangements P

_{1}must occur whereas P_{4}and P_{5}do not occur.Here, we Should choose P

_{1}every time, so now, we choose 4 persons out of 9 persons.From that 9 persons, we don’t need to choose P

_{4}and P_{5}, so This can be done in ways.Therefore, number of selections =

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

Total number of ways =

â‡’ 4200.

Therefore, number of such possible arrangements is 4200.

**Question 3: How many words, with or without meaning can be formed from the letters of the word “MONDAY” assuming that no letter is repeated. if**

**(i) 4 letters are used at a time.**

**Solution:**

Given, six letter word “MONDAY”.

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

These four letters can be arranged in 4! ways.

Therefore, total number of ways =

â‡’ 360.

**(ii) All letters are used at a time.**

**Solution:**

Given, six letter word “MONDAY”.

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

These 6 letters can be arranged in 6! ways.

Therefore, Total number of ways =

â‡’ 1 Ă— 720 = 720

**(iii) All letters are used but first letter is a vowel.**

**Solution:**

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 ways.

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

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

Therefore, total number of ways =

â‡’ 2\times1\times120=240.

**Question 4: Find the number of permutations of ****n**** distinct things taken ****r**** together, in which 3 particular things must occur together.**

**n**

**r**

**Solution:**

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 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 =

**Question 5: How many words of each 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE?**

**Solution:**

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

These five letters can be arranged in 5! ways.

Therefore, Total number of words formed =

â‡’ 4 Ă— 6 Ă— 120 = 2880

**Question 6: Find the number of permutations of ****n**** different things taken ****r**** at a time such that two specified things occur together?**

**n**

**r**

**Solution: **

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 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 =

â‡’

â‡’

**Question 7: Find the number of ways in which: (a) a selection (b) an arrangement, of four letters can be made from the letters of the word ‘PROPORTION’.**

**Solution: **

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.

- 3 alike letters and 1 distinct letter
- 2 alike letters and 2 distinct letters
- 2 alike letters of I kind and 2 alike letters of other kind.
- all distinct letters.
lets consider,

case-1: 3 alike and 1 distincthere we find that only one 3 alike letter in the word (‘OOO’).

So, choosing 1 distinct letter from remaining 5 distinct letters is

â‡’ 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

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

â‡’ 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

â‡’ 3 ways

Case-4: all distinct letters.Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is

â‡’ 15 ways.

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

â‡’

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 =

similarly, arrangement of 4 letters where 3 are alike is

total number of ways are

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

Total number of ways are

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

Total number of ways are

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 Ă— 4! = 360

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

â‡’

20+360+18+360 = 758 ways.

**Question 8: How many words can be formed by taking 4 letters at a time from the letters of word ‘MORADABAD’?**

**Solution: **

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.

- 3 alike letters and 1 distinct letter
- 2 alike letters and 2 distinct letters
- 2 alike letters of I kind and 2 alike letters of other kind.
- all distinct letters.
lets consider,

case-1: 3 alike and 1 distincthere we find that only one 3 alike letter in the word (‘AAA’).

So, choosing 1 distinct letter from remaining 5 distinct letters is

Number of ways of arranging them is

â‡’ 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

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

Number of ways of arranging them is

â‡’ 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

Number of ways of arranging them is

â‡’ ways

Case-4: all distinct letters.Here, we have 6 different letters, number of ways in choosing 4 letters out of 6 is

Number of ways of arranging them is 4!

â‡’ ways.

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

â‡’

20+240+6+360 = 626 ways.

**Question 9: A **businessman** hosts a dinner to 21 guests. He is having 2 round tables which can accommodate 15 and 6 persons each. In how many ways can he arrange the guests?**

**Solution:**

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 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 ways.

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

Total number of ways =

**Question 10: Find the number of combinations and permutations of 4 letters taken from the word ‘EXAMINATION’.**

**Solution:**

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.

- 2 alike letters and 2 distinct letters
- 2 alike letters of I kind and 2 alike letters of other kind.
- 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

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

Number of ways of arranging them is

â‡’ 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

Number of ways of arranging them is

â‡’ ways

Case-3: all distinct letters.Here, we have 6 different letters, number of ways in choosing 4 letters out of 8 is

Number of ways of arranging them is 4!

â‡’ ways.

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

â‡’

756+18+1680 = 2454 ways.

**Question 11: A tea party is arranged for 16 persons along two sides of a long table with 8 chairs on each side. Four persons wish to sit on one particular side and two on **another** side. In how many ways can they be seated?**

**Solution: **

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 ways.

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

This can be done in ways.

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

Therefore, total number of ways =

â‡’

## Please

Loginto comment...