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## Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 4 Combinatorics and Mathematical Induction Ex 4.5

Question 1.

The sum of the digits at the 10th place of all numbers formed with the help of 2, 4, 5, 7 taken all at a time is ……..

(a) 432

(b) 108

(c) 36

(d) 18

Solution:

(b) 108

Hint.

The given digits are 2, 4, 5 ,7

The unit place can be filled with the digit 2. Then the remaining Ten’s , hundred’s, Thousand’s place can be filled with the remaining three digits 4, 5, 7 in 3P_{3} ways.

∴ There will be 3 × 2 × 1 = 6 four digit numbers whose unit place is 2.

Similarly, there are 6 four digit numbers whose unit place is 4 , 6 four digit numbers whose unit place is 5 and 6 four digit numbers whose unit place is 7.

∴ Total in the unit place

= 6 × 2 + 6 × 4 + 6 × 5 + 6 × 7

= 6 × (2 + 4 + 5 + 7)

= 6 × 18 = 108

Sum of the digit at the unit place = 108

∴ Sum of the digit at the tens place = 108

Question 2.

In an examination, there are three multiple-choice questions and each question has 5 choices. A number of ways in which a student can fail to get all answer correct is ……..

(a) 125

(b) 124

(c) 64

(d) 63

Solution:

(b) 124

Hint.

Each question has 5 options in which 1 is correct.

So the number of ways of getting the correct answer for all three questions is 5^{3} = 125

So the number of ways in which a student can fail to get all answer correct is < 125 (i.e.) 125 – 1 = 124

Question 3.

The number of ways in which the following prize be given to a class of 30 boys first and second in mathematics, first and second in physics, first in chemistry and first in English is ……….

(a) 30^{4} × 29^{2}

(b) 30^{3} × 29^{3}

(c) 30^{2} × 29^{4}

(d) 30 × 29^{5}

Solution:

(a) 30^{4} × 29^{2}

Hint.

Number of students = 30

First prize can be given to any one of 30 students

= 30 × 30 × 30 × 30 = 30^{4} ways

Second prize can be given to anyone of the remaining 29 students = 29 × 29 = 29^{2} ways

∴ Total number of ways prizes can be given = 30^{4} × 29^{2} ways

Question 4.

The number of 5 digit numbers all digits of which are odd is ………

(a) 25

(b) 5^{5}

(c) 5^{6}

(d) 625

Solution:

(b) 5^{5}

Hint. The odd number are 1, 3, 5, 7, 9

Number of odd numbers = 5

We need a five-digit number So the number of five-digit number = 5^{5}

Question 5.

In 3 fingers, the number of ways four rings can be worn is …… ways.

(a) 4^{3} – 1

(b) 3^{4}

(c) 68

(d) 64

Solution:

(b) 3^{4}

Hint.

Number of rings = 4

Each finger can be worn rings in 4 ways.

∴ Number of ways of wearing four rings in three fingers

= 4 × 4 × 4

= 64

Question 6.

(a) 7 and 11

(b) 6 and 7

(c) 2 and 11

(d) 2 and 6

Solution:

(b) 6 and 7

Question 7.

The product of r consecutive positive integers is divisible by ………

(a) r!

(b) (r – 1)!

(c) (r + 1)!

(d) r!

Solution:

(a) r!

Hint.

The product of r consecutive positive integers is

1 × 2 × 3 × …………… × r = r!

which is divisible by r!

Also, 1 × 2 × 3 × …………… × r = r!

= ( r – 1) ! × r

which is divisible by (r – 1) !

Question 8.

The number of 5 digit telephone numbers which have none of their digits repeated is

(a) 90000

(b) 10000

(c) 30240

(d) 69760

Solution:

(d) 69760

Hint.

The number of 5 digit telephone numbers which have none of their digits repeated is ^{10}P_{5} = 30240

Thus the required telephone number is 10^{5} – 30240 = 69760

Question 9.

If a^{2} – ^{a}C_{2} = a^{2} – ^{a}C_{4} then the value of ‘a’ is ….

(a) 2

(b) 3

(c) 4

(d) 5

Solution:

(b) 3

Hint.

a_{2} – a = 2 + 4 = 6

a_{2} – a – 6 = 0

(a – 3) (a + 2) = 0 ⇒ a = 3

Question 10.

There are 10 points in a plane and 4 of them are collinear. The number of straight lines joining any two points is ……..

(a) 45

(b) 40

(c) 39

(d) 38

Solution:

(b) 40

Hint.

Question 11.

The number of ways in which a host lady invite 8 people for a party of 8 out of 12 people of whom two do not want to attend the party together is

(a) 2 × ^{11}C_{7} + ^{10}C_{8}

(b) ^{11}C_{7} + ^{10}C_{8}

(c) ^{12}C_{8} – ^{10}C_{6}

(d) ^{10}C_{6} + 2!

Solution:

(c) ^{12}C_{8} – ^{10}C_{6}

Hint.

Number of the way of selecting 8 people from 12 in ^{12}C_{8}

∴ out of the remaining people, 8 can attend in ^{10}C_{8}

The number of ways in which two of them do not attend together = ^{12}C_{8} – ^{10}C_{6}

Question 12.

The number of parallelograms that can be formed from a set of four parallel lines intersecting another set of three parallel lines …….

(a) 6

(b) 9

(c) 12

(d) 18

Solution:

(d) 18

Hint.

Number of parallelograms = ^{4}C_{2} × ^{3}C_{2 }= 6 × 3 = 18

Question 13.

Everybody in a room shakes hands with everybody else. The total number of shake hands is 66. The number of persons in the room is …….

(a) 11

(b) 12

(c) 10

(d) 6

Solution:

(b) 12

Hint.

Question 14.

The number of sides of a polygon having 44 diagonals is ……….

(a) 4

(b) 4!

(c) 11

(d) 22

Solution:

(c) 11

Hint:

Question 15.

If 10 lines are drawn in a plane such that no two of them are parallel and no three are concurrent, then the total number of points of intersection are ………

(a) 45

(b) 40

(c) 10!

(d) 2^{10}

Solution:

(a) 45

Hint:

Question 16.

In a plane there are 10 points are there out of which 4 points are collinear, then the number of triangles formed is …….

(a) 110

(b) ^{10}C_{3}

(c) 120

(d) 116

Solution:

(d) 116

Hint:

Question 17.

In ^{2n}C_{3} : ^{n}C_{3} = 11 : 1 then n is ………

(a) 5

(b) 6

(c) 11

(d) 1

Solution:

(b) 6

Hint.

Question 18.

^{(n – 1)}C_{r} + ^{(n – 1)}C_{(r – 1)} is ………

(a) ^{(n + 1)}C_{r}

(b) ^{(n – 1)}C_{r}

(c) ^{n}C_{r}

(d) ^{n}C_{r – 1}

Solution:

(c) ^{n}C_{r}

Question 19.

The number of ways of choosing 5 cards out of a deck of 52 cards which include at least one king is …….

(a) ^{52}C_{5}

(b) ^{48}C_{5}

(c) ^{52}C_{5} + ^{48}C_{5}

(d) ^{52}C_{5} – ^{48}C_{5}

Solution:

(d) ^{52}C_{5} – ^{48}C_{5}

Hint.

Selecting 5 from 52 cards = ^{52}C_{5}

selecting 5 from the (non-king cards 48) = ^{48}C_{5}

∴ Number of ways is ^{52}C_{5} – ^{48}C_{5}

Question 20.

The number of rectangles that a chessboard has ……

(a) 81

(b) 99

(c) 1296

(d) 6561

Solution:

(c) 1296

Hint. Number of horizontal times = 9

Number of vertical times = 9

Selecting 2 from 9 horizontal lines = ^{9}C_{2}

Selecting 2 from 9 vertical lines = ^{9}C_{2}

Question 21.

The number of 10 digit number that can be written by using the digits 2 and 3 is ……..

(a) ^{10}C_{2} + ^{9}C_{2}

(b) 2^{10}

(c) 2^{10} – 2

(d) 10!

Solution:

(b) 2^{10}

Hint.

Selecting the number from (2 and 3)

For till the first digit can be done in 2 ways

For till the second digit can be done in 2 ways ….

For till the tenth digit can be done in 2 ways

So, total number of ways in 10 digit number = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2^{10}

Question 22.

If P_{r} stands for ^{r}P_{r} then the sum of the series 1 + P_{1} + 2P_{2} + 3P_{3} + … + nP_{n} is ……..

(a) P_{n + 1}

(b) P_{n + 1} – 1

(c) P_{n – 1} + 1

(d) ^{(n + 1)}P_{(n – 1)}

Solution:

(b) P_{n + 1} – 1

Hint:

1 + 1! + 2! + 3! + … + n!

Now 1 + 1 (1!) = 2 = (1 + 1)!

1 + 1 (1!) + 2(2!) = 1 + 1 + 4 = 6 = 3!

1 + 1(1!) + 2(2!)+ 3(3!) = 1 + 1 + 4 + 18 = 24 = 4!

1 + 1(1!) + 2(2!) + 3(3!) ….+ n(n!) = (n + 1) ! – 1

= ^{n + 1}P_{n + 1} – 1 = P_{n + 1} – 1

Question 23.

The product of first n odd natural numbers equals …….

Solution:

Hint:

Question 24.

If ^{n}C_{4}, ^{n}C_{5}, ^{n}C_{6} are in AP the value of n can be ………..

(a) 14

(b) 11

(c) 9

(d) 5

Solution:

(a) 14

Hint:

30 + n^{2} – 9n + 20 – 12n + 48 = 0

n^{2} – 21 n + 98 = 0

(n – 7) (n – 14) = 0

n = 7 (or) 14

Question 25.

1 + 3 + 5 + 7 + + 17 is equal to ………

(a) 101

(b) 81

(c) 71

(d) 61

Solution:

(b) 81

Hint: