Important Questions for CBSE Class 11 Maths Chapter 7 – Permutations and Combinations -CoolGyan

CBSE Class 11 Maths Chapter-7 Important Questions - Free PDF Download

Free PDF download of Important Questions with solutions for CBSE Class 11 Maths Chapter 7 - Permutations and Combinations prepared by expert Maths teachers from latest edition of CBSE(NCERT) books. CoolGyan.Org to score more marks in your Examination.

1 Marks Questions

1. Evaluate

Ans.

2.If find

Ans.

3.The value of 0! Is?

Ans.

4.Given 5 flags of different colours here many different signals can be generated if each signal requires the use of 2 flags. One below the other

Ans.First flag can be chosen is 5 ways

Second flag can be chosen is 4 ways

By total number of ways

5.How many 4 letter code can be formed using the first 10 letter of the English alphabet, if no letter can be repeated?

Ans.First letter can be used in 10 ways

Second letter can be used in 9 ways

Third letter can be used in 8 ways

Forth letter can be used in 7 ways

By total no. of ways

6.A coin is tossed 3 times and the outcomes are recorded. How many possible out comes are there?

Ans.Total no. of possible out comes

7.Compute

Ans.

8.If

Ans.Given

9.In how many ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colours.

Ans.No. of ways of selecting 9 balls

10.Find if

Ans.

Rejected. Because if we put the no. in the factorial is –ve.

11.If find

Ans.

12.Write relation between and

Ans.

13.What is

Ans.Is multiplication of consecutive natural number

14.If what is the value of

Ans. then

15.How many words, with or with not meaning each of 2 vowels and 3 consonants can be flamed from the letter of the word DAUGHTER?

Ans.In the word DAUGHTER There are 3 vowels and 5 consonants out of 3 vowels, 2 vowels can be selected in ways and 3 consonants can be selected in ways 5 letters 2 vowel and 3 consonant can be arranged in 5! Ways

Total no. of words

16.Convert the following products into factorials

Ans.

17.Evaluate

Ans.

18.Evaluate

Ans.

19.What is the value of

Ans.

20.Find if

Ans.Given

21.Evaluate

Ans.

22.If then what is the value of

Ans.

23.How many 3 digit numbers can be formed by using the digits 1 to 9 if no. digit is repeated

Ans.

24.Convert into factorial 2.4.6.8.10.12

Ans.

25.How many words with or without meaning can be formed using all the letters of the word ‘EQUATION’ at a time so that vowels and consonants occur together

Ans. In the word ‘EQUATION’ there are 5 vowels [A.E.I.O.U.] and 3 consonants [Q.T.N]

Total no. of letters = 8

Arrangement of 5 vowels =

Arrangements of 3 consonants =

Arrangements of vowels and consonants =

Total number of words

4 Marks Questions

1. How many words, with or without meaning can be made from the letters of the word MONDAY. Assuming that no. letter is repeated, it

(i) 4 letters are used at a time

(ii) All letters are used but first letter is a vowel?

Ans. Part-I In the word MONDAY there are 6 letters

4 letters are used at a time

Total number of words

Part-II All letters are used at a time but first letter is a vowel then OAMNDY

2 vowels can be arranged in 2! Ways

4 consonants can be arranged in 4! Ways

Total number of words

2.Prove that

Ans.Proof L.H.S.

3. A bag contains 5 black and 6 red balls determine the number of ways in which 2 black and 3 red balls can be selected.

Ans. No. of black balls =5

No. of red balls = 6

No. of selecting black balls = 2

No. of selecting red balls = 3

Total no. of selection

4.In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

Ans.Let us first seat 0 the 5 girls. This can be done in 5! Ways

X G X G X G X G X G X

There are 6 cross marked placer and the three boys can be seated in ways

Hence by multiplication principle

The total number of ways

5. How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE

Ans.In the INVOLUTE there are 4 vowels, namely I.O.E.U and 4 consonants namely M.V.L and T

The number of ways of selecting 3 vowels

Out of

The number of ways of selecting 2 consonants

Out of

No of combinations of 3 vowels and 2 consonants

5 letters 2 vowel and 3 consonants can be arranged in 5! Ways

Therefore required no. of different words

6. Find the number of arrangements of the letters of the word INDEPENDENCE. In how many of these arrangements

(i) do the words start with P

(ii) do all the vowels always occur together

Ans. The number of letters in the word INDE PENDENCE are 12 In which E repeated in 4 times. N repeated 3 times. D repeated 2 times

(i) If the word starts with P The position of P is fined

Then the no. of arrangements

(ii)All the vowels always occur together There are 5 vowels in which 4 E’s and 1 I

NDPNDNC

Total letters are 8 letters Can be arranged in ways

Also 5 vowels can be arranged in ways.

required number of arrangements

7.Find

Ans.Given

8. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Ans. Given total no. of players =17

5 players can bowl.

no. of bat’s man =17-5=12

Out of 5 bowlers 4 can be choose in ways

Out of 12 bat’s man (11-4)=7 bat’s man can be choose in ways

Total no. of selection of 11 players

9. How many numbers greater than 1000000 can be formed by using the digits 1,2,0,2,4,2,4?

Ans.Given digits Are 7

The no. of arrangements of 7 digits

If 0 is in extreme left position.

The no. of arrangements of 6 digits

Required numbers

10. In how many ways can the letters of the word ASSA SS IN ATION be arranged so that all the S’s are together?

Ans.In the word ASSA SSI N ATION there are 13 letters

But all S are together. Then no. of letters 10 [4 S take 1]then required no. of arrangements

11. The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be flamed from the alphabet?

Ans. No. of vowels = 5

No. of consonants = 21

2 vowels can be selected in ways

2consonents can be selected in ways

No. of arrangements of 4 letters [2 vowel and 2 consonants] =4!

Required no. of wards

12. In how many of the distinct permutations of the letters in MISSISSIPPI do the four I’s not come together?

Ans.In the word MISSISSIPPI no. of letters = 11

In which 4 I’s, 4 S’s and 2P

Total no. of words

When four I’s come together

Then four I’s as one letter

And other letters are 7

Then no. of words when four I’s come together

Then the no. of permutations when four I’s do not come together

13. In how many ways can 4 red, 3 yellow and 2 green discs be arranged in a row if the discs of the same colors are in distinguishable?

Ans.Total no. of discs

Out of 9 discs 4 are of some kind, 3are of same kind, 2 are of same kind

There for the number of arrangements

14. Find the number of permutations of the letters of the word ALLAHABAD.

Ans.In the word ALLAHABAD no. of letters = 9

In which four A’s and two L’s

The no. of permutations

15. How many 4 letter code can be formed using the first 10 letters of the English alphabet if no letter can be repeated?

Ans.There are 10 letters of English alphabet

For making 4 letter code

First letter can be choose in 10 ways

Second letter can be choose in 9 ways

Third letter can be choose in 8 ways

Forth letter can be choose in 7 ways

By fundamental principle of multiplication

Total no. of code

16. Determine the number of ways of choosing 5 cards out of a deck of 52 cards which include exactly one ace.

Ans.In a deck of 52 cards, there are 4 aces and 48 other cards. Here, we have to choose exactly one ace 4 other cards.

The number of ways of choosing one ace out of 4 aces

The number of ways of choosing 4 cards out of the other 48 cards

Corresponding to one way of choosing an ace, there are ways of choosing 4 other cards. But there are ways of choosing aces, there fore, the required number of ways

17.How many numbers greater than 56000 and farmed by using the digits 4,5,6,7,8, no digit being repeated in any number?

Ans.Number greater than 56000 and farmed by using the digits 4,5,6,7,8 are of types

Now numbers of type are

Number of type are

Hence required number of numbers is

18.Find if are in the ratio 2:1

Ans.Given

But, for are not meaningful, therefore,

19. Prove that

Ans.

20. How many 4 letter words with or without meaning, can be formed out of the letters of the word ‘LOGARITHMS’, if repetition of letters is not allowed?

Ans.There are 10 letters in the word ‘LOGARITHMS’

For making 4 letter word we take 4 at a time

No. of arrangements 10 letters taken 4 at a time

21. From a class of 25 students 10 are to be chosen for an excursion Party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can excursion party be chosen?

Ans. Total no. of students = 25

No. of students to be selected = 10

I case :

3 students all of them will join the excursion party.

Then remaining 7 students will be selected out of (25-3 = 22) in ways

II case :

All 3 students will not join the party then 10 students will be selected in ways

Total no. of selection

22. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each colour.

Ans. No. of red balls = 6

No. of white balls = 5

No. of blue balls = 5

No. of selecting each colour balls = 3

required no. of selection =

23. Find the number of 3 digit even number that can be made using the digits 1, 2, 3, 4, 5, 6, 7, if no digit is repeated?

Ans. For making 3 digit even numbers unit place of digit can be filled in 3 ways. Ten’s place of digit can be filled in 5 ways. Hundred place of digit can be filled in 4 ways. Required number of 3 digit even number

24. If find the values of

Ans. Given that

And

Solving eq. and

25. Prove that the product of consecutive positive integer is divisible by

Ans. Suppose & consecutive positive integers are

Then product

which is divisible by

6 Marks Questions

1. A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has :

(i) no girl? (ii) at least one boy and one girl? (iii) at least 3 girls?

Ans. Number of girls = 4

Number of boys = 7

Number of selection of members = 5

(i) If team has no girl

We select 5 boys

Number of selection of 5 members

(ii) At least one boy and one girl the team consist of

Boy	Girls
1	4
2	3
3	2
4	1

The required number of ways

(iii) At least 3 girls

Girls	Boys
3	2
4	1

The required number of ways

2. Find the number of words with or without meaning which can be made using all the letters of the word. AGAIN. If these words are writer as in a dictionary, what will be the 50^th word?

Ans. In the word ‘AGAIN’ there are 5 letters in which 2 letters (A) are repeated

Therefore total no. of words

If these words are written as in a dictionary the number of words starting with Letter A. [A A G I N]

The no. of wards starting with G [G A A I N]

The no. of words starting with I [I A A G N]

Now

Total words

Words = N A A G I

Words = N A A I G

3. What is the number of ways of choosing 4 cards from a pack of 52 playing cards? In how many of there

(i) Four cards one of the same suit

(ii) Four cards belong to four different suits

(iii) Are face cards.

(iv) Two are red cards & two are black cards.

(v) Cards are of the same colour?

Ans.The no. of ways of choosing 4 cards form 52 playing cards.

(i) If 4 cards are of the same suit there are 4type of suits. [diamond club, spade and heart] 4 cards of each suit can be selected in ways

Required no. of selection

(ii) If 4 cards belong to four different suits then each suit can be selected in ways required no. of selection

(iii)If all 4 cards are face cards. Out of 12 face cards 4 cards can be selected in ways.

required no. of selection

(iv) If 2 cards are red and 2 are black then. Out of 26 red card 2 cards can be selected in ways similarly 2 black card can be selected in ways

required no. of selection

(v) If 4 cards are of the same colour each colour can be selected in ways

Then required no. of selection

4. If and find the value of and

Ans. Given that

Solving eq and eq we get and

5. Find the value of such that

Ans (i)

Is rejected

Because negative factorial is not defined

(ii)

6. A committee of 7 has to be farmed from 9 boys and 4 girls in how many ways can this be done when the committee consists of

(i) Exactly 3 girls?

(ii) Attest 3 girls?

(iii) Atmost 3 girls?

Ans. No. of boys = 9

No. of girls = 4

But committee has 7 members

(i) When committee consists of exactly 3 girls

	Boys	Girls
	9	4
Selecting member	4	3	=7

Required no. of selection

(ii) Attest 3 girls

	Boys	Girls
	9	4
Selecting member	4	3	=7
	3	4	=7

The required no. of selections

(iii) Atmost 3 girls

	Boys	Girls
	9	4
	7	0	=7
Selecting member	6	1	=7
	5	2	=7
	4	3	=7

Then required no. of selection

7. In how many ways can final eleven be selected from 15 cricket players’ if

(i) there is no restriction

(ii) one of then must be included

(iii) one of them, who is in bad form, must always be excluded

(iv) Two of then being leg spinners, one and only one leg spinner must be included?

Ans. (i) 11 players can be selected out of 15 in ways

ways ways=1365 ways

(ii) Since a particular player must be included, we have to select 10 more out of remaining 14 players.

This can be done in ways ways

ways = 1001 ways

(iii) Since a particular player must be always excluded, we have to choose 11 ways out of remaining 14

This can be done in ways ways

ways ways.

(iv) One leg spinner can be chosen out of 2 in ways. Then we have to select 10 more players get of 13 (because second leg spinner can’t be included). This can be done in ways of choosing 10 players. But these are ways of choosing a leg spinner, there fore, by multiplication principle of counting the required number of ways

8. How many four letter words can be formed using the letters of the letters of the word ‘FAILURE’ so that

(i) F is included in each word

(ii) F is excluded in each word.

Ans.There are 7 letters in the word ‘FAILURE’

(i) F is included for making each word using 4 letters

F is already selected

Then other 3 letters can be selected out of 6 are ways

Also arrangements of 4 letters are 4! Ways so:

Total no. of words

(ii) F is excluded in each word

Out of 6 letters are choose 4 letters in ways

Also arrangement of 4 letters are 4! Ways so:

Total no. of words

9. A committee of 5 is to be formed out of 6 gents and 4 Ladies. In how many ways this can be done, when

(i) at least two ladies are included?

(ii) at most two ladies are included?

Ans. No of person to form committee = 5

No. of gents = 6 and No. of ladies = 4

(i) At least two ladies are included

Ladies [4] Gents [6]

Either we Select 2 and 3

3 and 2

4 and 1

required number of selection

(ii) At most two ladies are included?

Ladies [4] Gents [6]

Either we select 0 and 5

Or 1 and 4

Or 2 and 3

required no. of selection.

10. In how many ways can the letters of the word PERMUTATIONS be arranged if the

(i) words start with P and with S

(ii) vowels are all together

(iii) There are always 4 letters between P and S?

Ans. In the word PERMUTATIONS there are 12 letters

(i) If the word start with P and end with S then position of P and S will be fixed. Then other 10 letters can be arranged in ways. But T occurs twice.