Students can Download Basic Maths Question Bank Chapter 2 Permutations and Combinations Questions and Answers, Notes Pdf, 2nd PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Basic Maths Question Bank Chapter 2 Permutations and Combinations

### 2nd PUC Basic Maths Permutations and Combinations One Mark Questions and Answers

Question 1.

Find the value of ^{8}P_{3
}Answer:

^{8}P_{3} = 8x 7 x6 = 336

Question 2.

Find the value of ^{14}C_{12
}Answer:

Question 3.

Out of 16 men, in how many ways a group of 7 men may be selected.

Answer:

7 men out of 16 men can be selected in ^{l6}C_{7} ways

Question 4.

In how many ways 5 boys can form a ring, The no of circular permatations

Answer:

(5-1)! 4!

Question 5.

In how many ways can 6 people stand in a Queue

Answer:

5 people can stand in a Queue in 5! ways = 120 ways

Question 6.

In how many ways can 9 different precious stones be set to form a necklace?

Answer:

The number of ways = \(\frac{(n-1) !}{2}=\frac{81}{2}\)

Question 7.

If ^{n}C_{4} = ^{n}C_{6} find n

Answer:

4 = n – 6, n = 10

∴ ^{n}C_{r} = ^{n}C_{nr}

Question 8.

If ^{n}P_{2} = 30 find n

Answer:

n(n – 1) = 30 ⇒ 6 x 5 = 30

∴ n = 6

Question 9.

If ^{n}C_{2} = 105 find n

Answer:

Question 10.

If ^{n}P_{4} = 360 find n

Answer:

\(\frac{9 !}{2 ! \times 2 ! \times 2 !}\)

n(n -1) (n – 2) (n – 4) = 360

⇒ 6 x 5 x 4 x 3 = 360 ⇒ n = 6

Question 11.

If ^{n}P_{3} = 24 find n

Answer:

\(\frac{9 !}{2 ! \times 2 ! \times 2 !}\)

n(n – 1 )(n – 2) = 24

⇒ 4 x 3 x 2 = 24

∴ n = 4

Question 12.

In how many ways can a candidate select 5 Questions out of 10 Questions.

Answer:

A candidate can select 5 Questions out of 10 Questions in ^{10}C_{5} ways.

Question 13.

In how many ways can 8 flowers of different colours be strung into a garland

Answer:

\(\frac{(n-1) !}{2}=\frac{7 !}{2}\)

Question 14.

Find the number of words formed using the letters of the following words.

Answer:

(i) “CRICKET”

(ii)”COOKS”

(iii) “GOOD”

(iv)”MISSION”

Question 15.

How many of the arrangements of the word “ABACUS” begin with A ?

Answer:

Since beginning Letter A is fixed, the remaining 5 letters can be permuted in 5! ways.

Question 16.

How many of the arrangements of the word “CRICKET” begin end with C

Answer:

Since the 1st & last letter should be C, the remaining 5 places can be aranged in 5! ways.

Question 17.

If ^{n}P. = 24 ^{n}C_{4} find the value of n

Answer:

n – 4 = 1

⇒ n = 1 + 4 = 5

Question 18.

Find the number of words that can be formed using the letters of the word “DELHT” beginning with D.

Answer:

D is fixed, the remaining 4 letters can be arranged in 4! = 24 ways.

Question 19.

Find the number of diagonals of a polygon of sides 8

Answer:

No of diagonals =

Question 20.

There are 14 points in a plane, out of which 4 are collinear. How many straight lines can be drawn?

Answer:

^{14}C_{2}–^{4}C_{2}-+1 St lines

Question 21.

How many Six digit numbers can be formed with the digits 2,7,6,1,9,8

Answer:

There are 6 digits, that can be formed in 6! ways = 72 ways.

Question 22.

In how many ways can 4 people occupy 6 vacant chains.

Answer:

4 people occupy 6 vacant chains in ^{6}P_{4} ways = 360 ways.

Question 23.

There are 4 routes to go from A to B, & 3 routes to go from B to C. In how many ways can you go from A to C via B.

Answer:

No of ways of go from A to C via B = 4 x 3 = 12 ways.

Question 24.

In how many ways can you go from A to B & return to A from a different route, if there are 10 ways of travelling from A to B

Answer:

No of ways of travelling from A to B = ^{10}P_{2} ways = 10 x 9 = 90 ways.

Question 25.

Find x if ^{9}C_{x }+ ^{9}C_{7} = ^{10}C_{7
}Answer:

^{n}C_{r} + ^{n}C_{r – 1 }= (n + 1)C_{r
}=7 – 1 = 6

x = 6

Question 26.

How many strainght lines can be formed from 10 points if none them are collinear.

Answer:

No of strainght lines that can be formed = ^{10}C_{2}

Question 27.

How many triangles can be formed from 8 non collinear points.

Answer:

Number of triangles that can be formed = 8C_{2}

Question 28.

If ^{n}C_{10} = ^{n}C_{15} find n

Answer:

n= 10+ 15 = 25

### 2nd PUC Basic Maths Permutations and Combinations Two Marks Questions and Answers

Question 29.

If ^{n}P_{4} =20 ^{n}P_{3}, find n

Answer:

Question 30.

In how many ways can 7 persons be seatea in a row If two persons always occupy the end seats.

Answer:

Last 2 places are fixed the remaining 5 persons can be arranged in 5! ways & those 2 persons can be arraged in 2! ways

Total arrangement = 5! x 2! = 240 ways.

Question 31.

In how many ways the word “CARROM” be arranged such that the 2R‘s are always Together.

Answer:

The word has 6 letters. Two R’s are always together.

∴ they can be taken as 1 unit (4+1= 5)

∴ the number of permutations = 5! = 120 ways.

Question 32.

If ^{n}P_{3} :^{n}P_{2} = 3:1 find n

Answer:

Question 33.

How many 5 digit numbers ending with 7 can be formed using the digits 1,2,5,7 digits can not be repeated.

Answer:

O cannot be in the 1st place

∴ 1st place can be filled in 4 ways &

remaining 4 places can be done in 4! ways.

∴ Total numbers of permatations = 4! x 4 = 96 ways

Question 34.

How many 3 digit numbers ending with 7 can be formed using the digits 1, 2, 5, 7, 8 digits can not be repeated.

Answer:

the last digit is fixed as 7 and remaining

2 places can be filled in by 4 digits in ^{4}P_{2} ways.

Question 35.

In how many ways can the letters of the word “HOPPER” be arranged.

Answer:

No of permatations = \(\frac{6 !}{2 !}\) = 360 ways

Question 36.

How many 6 digit numbers can be formed from the digits 1,2,3,4,5,6 (No digit being repeated) which are divisable by 5

Answer:

5 is to be fixed at the end

∴ the numbers of permutations = 5! ways = 120 ways

Question 37.

If ^{n}C^{8} = ^{n}C_{n} find ^{n}C_{5
}Answer:

^{n}C_{8} = ^{n}C_{n-12
}n -12 = 8 ⇒ n = 20

^{n}C_{5} = ^{20}C_{5}= 15504

Question 38.

If ^{n}P_{r} = 30240, ^{n}C_{r}= 252 find r

Answer:

Question 39.

Find the number of ways in which a committee of 4 students and 2 lecturers can be chosen out of 10 students and lecturers.

Answer:

No of selections of 4 students out of 10 students and 2 lecturers out 8 lecturers in

^{l0}C_{4} x ^{8}C_{2} ways = 5880 ways.

Question 40.

Find r if ^{25}C_{f+3} = ^{2}^{s}C_{2r}

Answer:

r + 3 = 2r – 3 ⇒ b = r or r =6

Question 41.

Find the number of straight lines and triangles that can be formed out of 20 points of which 8 are collinear

Answer:

No of straight lines = ^{20}C_{2} – ^{8}C_{2} +1

= 190 – 28+1

= 163 lines

No of triangles that can be formed = ^{20}C_{3} – ^{S}C_{3
}1140 – 56= 1084

Question 42.

Find the number of paralleograms that can be formed from a set of 6 parallel lines intersecting another set of 4 parallel lines

Answer:

Number of parallelograms = ^{m}C_{2} x ^{n}C_{2} = ^{6}C_{2} x ^{4}C_{2}

= 15 x 6 = 90

Question 43.

Find the number of diagonals of a polygon of 20 sides

Answer:

Numbers of diagonals = ^{n}C_{2 }– n

= ^{20}C_{2} – 20 = 170

Question 44.

In a party each person shakes hands with every one else. If there are 25 members in the party calculate the numbers of handshakes

Answer:

Number of handshakes of 25 members

= ^{25}C_{2} = 300

Question 45.

In how many ways can 6 people be chosen out of 10 people of one particular person is always included.

Answer:

6 people are to be chosen out of 10 people of one person is always included

= ^{9}C_{5} = 126 ways

Question 46.

If ^{n}P_{r} = 240 & ^{n}C_{r} = 120 find n and r

Answer:

Question 47.

How many 5 digit numbers can be formed with the digits 2,3,5,7,9 which lies between 30,000 and 90,000

Answer:

The 1st place van be filled in 4 ways us there are 4 numbers greater than 2 since one digital is already filled ,the 2nd place can be filled in 4 ways by the remanining 3 places can be filled in 3, 2, 1 ways.

∴ total permutation = 4 x 4 x 3 x 2 = 96 ways

Question 48.

In how many ways 3 boys and 5 girls can be arranged in a row so that no 2 boys are together.

Answer:

3 boys have to be Seated among 5 girls, there are 6 places 3 boys can be arranged in ^{6}P_{3} ways & 5 girls can be seated in 5! ways.

∴ 5! ^{6}P_{3}

Question 49.

In how many ways can the 7 colours of the rainbow be arranged so that the red and the blue colours are always together.

Answer:

Red & blue are together can be taken as 1 unit and this can be arranged in 6! ways Red & blue can be arranged in 2! ways.

∴ the number of ways = 6! x 2! = 1440

Question 50.

How many words can be formed from the letters of the word “DAUGHTER” So that the vowels do not appear together?

Answer:

The total 8 letters can be arranged in 8! ways let the 3 vowels (A, E, U) be together can be taken as one unit & remaining 5 letters & 1 unit of vowels can be done in 6! ways Again 3 vowels can permute in 3! ways.

∴ The number of ways in which the vowels are together = 6! 3!

∴ the total number of ways in which the vowels are not together = total no of ways – no of ways in which the vowels are together = 8! – 6! 3!

### 2nd PUC Basic Maths Permutations and Combinations Three or Four Marks Questions and Answers

Question 51.

In how many ways can 7 students & 4 Teachers be seated in row such that no two teachers are together

Answer:

There are 8 places between 7 students & 4 teachers can be arranged in 8 places in ^{8}P_{4} ways & 7 students can be arranged in 7! ways

∴ The total number of ways = 7! x ^{8}P_{4}

Question 52.

In how many ways can 7 books of arts, 5 books of fiction, 4 books on adventure be arranged so that books of the same category are always together.

Answer:

First consider 7 books as 1 unit, 5 books as 1 unit & 4 books as 1 unit, Total of 3 unit can be arranged in 3! & arts books in 7!, fiction books in 5! adventure books as 4! ways .’. Total number of ways = 3! x 5! x 7! x 4!

Question 53.

5 men and 4 women are to be seated in a row so that women occupy the even places. How many such arrangement are possible.

Answer:

There are 5 men & 4 women, 5 men be arranged in 5! ways there are 4 even places between 5 men this can be done in 4! ways.

∴ No of permutations = 5! x 4! = 2880 ways.

Question 54.

How many 5 digit even numbers can be formed using the digits 0,1,2,3,4, 5,

Answer:

O cannot be fixed in the I place. In the units place (0 or 2 or 4) can be filled in 3 ways I place can be filled in 5 ways the remainging places can be filled in 6 x 6 x 6 ways.

∴ the numbers of permatations = 3 x 5 x 6 x 6 x 6 = 3240 ways.

Question 55.

In how many ways the letters of the word “ACCOUNTANT” be arrarged? In how many of then vowels are always together.

Answer:

The word ACCOUNTANT has 10 letters this can be done in \(\frac{10 !}{2 ! \times 2 ! \times 2 ! \times 2 !}\)

The vowels are { A, O, U, A} is taken as one unit + 6 letters = 7 letters & 4 vowels can be arranged in 4! ways.

∴ The number of permutations \(=\frac{7 ! \times 4 !}{2 ! \times 2 ! \times 2 ! \times 2 !}\)

Question 56.

Find the number of permatations of the letters of the word “MISSI SSI PPI” In how many of these (i) 4S^{3} are together (ii) The 4S^{3} are not together (iii) begin with MISS (IV) begin with SIP

Answer:

Total number of permatations = \(\frac{11 !}{4 ! \times 4 ! \times 2 !}\)

S = 1

I = 4

P = 2

(i) 4S^{3} are together consider 4S^{3} are 1 unit =1+7 = 8

This can be arranged = \(\frac{8 !}{4 ! \times 2 !} \times \frac{4 !}{4 !}\)

\(=\frac{8 !}{4 ! \times 2 !}=840\)

(ii) 4S^{3} are not together = Total – Together \(=\frac{11 !}{4 ! \times 4 ! \times 2 !}-\frac{8 !}{4 ! \times 2 !}\)

(iii) Begin with MISS

MISS is Fixed & the remaining places can be filled in \(\frac{7 !}{3 ! \times 2 ! \times 2 !}\)

(iv) Begin with SIP

Fix SIP & the remaining places can be filled in \(\frac{8 !}{3 ! \cdot 3 !}\)

Question 57.

Find the numbers of permatation of the letters of the word “COMMITIEE”.

(a) How many of them begin with T & end with T

(b) In how many ways all the vowels are together

(c) In how many ways no two vowels are together

(d) In how many of them end with MITE

Answer:

The word has 9 letters of which M – 2, T – 2, E – 2

Total numbers of permatations \(=\frac{9 !}{2 ! \cdot 2 ! \cdot 2 !}\)

(i) Begin with T & end with T

Fix T in the begining & in the end the remaining letters can be arranged is \(\frac{7 !}{2 ! 2 !}\)

(ii) All vowels are together

Vowels are ( O, I, E, E) as 1 unit The remaining letters 5 + 1 = 6 can be arranged in

\(=\frac{6 !}{2 ! .2 !} \times \frac{4 !}{2 !}=\frac{720 \times 24}{8}=2160\)

(iii) No Two vowels are together,

First arrange the five consonents C M M T T there are 6 places, these 5 consonents can be \(\frac{5 !}{2 ! \times 2 !}\) and there are 4 vowels and 6 gaps this can be done in \(\frac{^{6} \mathrm{P}_{4}}{2 !}\)

∴ No of permatations \(\frac{5 !}{2 ! \times 2 !} \times \frac{^{6} P_{4}}{2 !}\)

(iv) End with MITE

Fix MIT E in the last 4 places

The remaining 5 letters can be arranged in 5! ways.

Question 58.

Find the number of permatations of the letters of the word ASSA SSINATION in how many of there.

(a) the vowels are in even places

(b) vowels are in odd places

(c) the word NATION is always present together

(d) Begins with ‘AS’ and ends with ‘AS’.

Answer:

The word has 13 letters of which A -3, S – 4, I – 2, N – 2

Total number of permatations = \(\frac{13 !}{3 ! \times 4 ! \times 2 ! \times 2 !}\)

(a) vowels are in even places

There are 6 vowels ( A, A, A, I, I, O) can be arranged in even places in \( \frac{6 !}{3 ! \times 2 !} ways\)

There are 7 consonents can be arranged in \(\frac{6 !}{3 ! \times 2 !} \times \frac{7 !}{4 ! \times 2 !}\)

∴ Total numbers of permatations = \(\frac{^{7} P_{6}}{3 ! \times 2 !} \times \frac{7 !}{4 ! \times 2 !}\)

(b) Vowels are in odd places

There are 7 odd places and the vowels can be arranged in odd places in \(\frac{^{7} P_{6}}{3 ! \times 2 !}\)

∴ Total no of permatations = \(\frac{^{7} P_{6}}{3 ! \times 2 !} \times \frac{7 !}{4 ! \times 2 !}\)

(c) The word NATION is always present together. The word NATION should be taken as 1 unit (NATION) + ASSSSAIT =1+7 = 8

∴ The No of permatations = \(\frac{8 !}{4 ! \times 2 !}\)

Begin with AS and end with AS fix AS in the I two places & Fix AS in the last 2 places.

The remaining 9 letters can be arranged in \(\frac{9 !}{2 ! \times 2 ! \times 2 !}\)

Question 59.

In how many ways can 7 english books, 8 Kannada books and 5 Hind, books be arranged in a shelf. In how many of these

(a) All Kannada books are together

(b) No two English books are together

(c) Books of the same Language are together

(d) All Kannada books are together & all Hindi books are together.

Answer:

Total No of Permatations = (7+8+5)! =20!

(a) Consider all Kannada books as 1 unit

(1+7+5 = 13) can be arranged in 13! & kannada books can be arranged in 8! ways

∴ No of permatations =13! x 8!

(b) No 3 English books are together there are 13 books & 14 gaps

there 13 books can be arranged in 13! ways and 7 English books can be arranged in 14 places in ^{14}P_{7 }ways.

∴ Total numbers of ways = 13! x ^{14}P_{7}

(c) Books of the Same Language are together

∴ 1 unit of K + 1 unit of E + 1 unit of 1 + = 3units can be arranged in 7! ways and Engilsh books in 8! and Hindhi books in 5! ways – Numbers of permatations

= 3! x 8! x7! x 5!

(d) Kannad books as 1 unit & Hindi books as 1 unit

∴ Total = 1+1+7 = 9 units can be done in 9! ways Kannada books & Hindi books then selves can be arranged in 8! x 5! ways

∴ The total no of permatations = 9! x 8! x 5!

Question 60.

A committee of 4 has to be chosen form 10 boys and 8 girls In how many can this be done if girls are in a majority.

Answer:

There are 10 boys & 8 girls of which 4 are selected in the following ways.

Question 61.

A team of 8 players has to be selected from 14 players. In how many ways the selections can be made if.

(a) Two particalar players are always included

(b) Two particular players are always excluded

Answer:

(a) there are 14 players of who 8 have to be selected 2 players are always

included = ^{12}C_{6} ways

(b) 2 players are always excluded the total no of selection = ^{12}C_{8} ways.

Question 62.

From a class of 12 boys and 10 girls, 10 students are to be chosen for a competition, including at least 4 boys and at least 4 girls the two girls who won the prize last year should be included in how many ways can the selections be made.

Answer:

10 students ere to be chosen for a competition out of 12 boys & 10 girls if 2 girls are to be included.

Question 63.

A box contains 5 red, 4 black and 3 white balls. How many selections of 8 balls can be made if the selection contains

(i) exactly 4 red, 2 black and 2 white balls (ii) atleast 3 red, atleast 3 black & atleast 1 white balls.

Answer:

(i) Exactly 4 red from 5, 2 black from 4 and 2 white from 3 balls can be selected in

^{5}C_{4} x ^{4}C_{2} x ^{3}C_{2} = 90 ways

(ii) Atleast 3 red, atleast 3 black and atleast 1 white ball be selected in the following ways.

Question 64.

An examination paper consists of 12 Questions divided in to parts A and B contains 7 Questions in part A and part B contains 5 Question A candidate is required to answer 8 Questions selceting at least 3 from each part. In how many ways can the candidate select the question.

Answer:

There are 12 Question divided in to 2 part Part A contians 7 question and part B contains 5 Questions 8 questions are to be answered by selecting atleast 3 from each part.

Question 65.

A man has 10 relatives, 4 of them are Ladies 3 gentlemen and 3 Children, In how many ways can he invite 7 relatives to dinner party so that

(i) there are exactly 2 ladies 3 gentlemen and 2 children

(ii) there are exactly 2 gentlemen & atleast 3 ladies

(iii) there are exactly 3 children, atleast Idady and atleast 2 gentlemen.

Answer:

There are 4 Ladies, 3 gentlemen and 3 Children, 7 persons should be invited to a party.

(i) No of ways of selection = ^{4}c_{2 }x ^{3}c_{3} x ^{3}c_{2 }= 18 ways

Question 66.

A committee of 5 are to be formed from 8 Americans and 5 Anglo Indians. In how many

ways can this be done when the committe contains

(i) Exactly two Anglo – Indians

(ii) At least two Anglo – Indians

Answer:

Exactly two Anglo – Indians can be selected in

(i) 8 American 5 Anglo Indians

3 – 2 = ^{4}C_{1} x ^{8}C_{5 }= 560 ways

Question 67.

A Team of eleven is to be chosen out of 16 cricket players of whom 4 are bowlers and 2 wicket- Keepers. In how many ways can the team be chosen so that (i) there are exactly 3 bowlers and 1 wicket-keeper? (ii) there are at least 3 bowlers & at least one wicket keeper.

Answer:

(i) 4 Bowlers 2 Wicket keepers 10 Others

Exactly 3 bowlers & 1 w k and 7 others can be selected in ^{4}C_{3} x ^{2}C_{1} x ^{10}C_{7
}= 960 ways

(ii) There are at least 3 bowlers & at least one wicket keeper can be selected an follows

Question 68.

A candidate is required to answer 6 out of 12 Questions which are divided into two groups containing 6 Questions in each group, Find the numbers of choices he has if he cannot attempt more than five Questions from any group.

A(6) B(6) No of ways

Answer:

Question 69.

From 4 officers and 8 Jawans. a committee of 6 is to be chosen to include exactly one officer find the numbers of such committes.

Answer:

Out of 4 officer, 10 officer is to be choosen & out of 7 Jawans we have to select 5 Jawans this can be done in ^{8}C_{5} ways.

Numbers of committe = ^{4}C_{1} x ^{8}C_{5
}= 224 ways

Question 70.

Find n if 2 ^{n}P = 2. ^{n}P

Answer:

2 n(2n – 1)(2n -2) = 2 .n (n – 1)(n – 2)(n – 3)

2(2n – 1)(n – 1) = (n – 1)(n – 2)(n – 3)

4n – 2 = n2 – 5n + 6

n2 – 9n + 8 = 0

(n – 8) (n – 1) = 0

n = 1 or n = 8

n ≠ 1 (cannot possible ∴ n ≥ r)

∴ n = 8