Chapter 7: Permutations and Combinations

(i)Thousand’s Place can be filled with 6 alone.

Hence, number of ways =1

Unit place can be filled with either 0 or 5.

Hence, number of ways =2

Hundred’s place can be filled with the remaining 8 digits.

Hence, the number of ways =8

Ten’s place can be filled with 7 digits.

Number of ways= 7.

Thus, the required number will be

=1 x 8 x 7 x 2

=112

We know that,

ⁿC_r

According to the question,

There are 10 people named P1, P2, P3, … P10.

Number of ways in which P₁ can be arranged =5! =120

Number of ways in which others can be arranged,

⁷C₄ = 7!/(4!3!) = 35

Therefore, the required number of arrangements =C₄⁷ × 5

=35 x 120

=4200

We know that,

ⁿC_r

We also know that,

According to the question,

Number of lamps in a hall =10

Given that,

One of the lamps can be switched on independently

Hence, the number of ways in which the hall can be illuminated is given by,

C₁¹⁰ + C₂¹⁰ + C₃¹⁰ + C₄¹⁰ + C₅¹⁰ + C₆¹⁰ + C₇¹⁰ + C₈¹⁰ + C₉¹⁰ + C₁₀¹⁰

=2¹⁰-1

=1024-1

=1023

We know that,

ⁿC_r

Drawing 1 black and 2 other ball = ³C₁ × ⁶C₂

Drawing 2 black and 1 other ball = ³C₂ × ⁶C₁

Drawing 3 black balls = ³C₃

Number of ways in which at least one black ball can be drawn =

=(1 black and 2 other )or( 2 black and 1 other )or (3 black)

³C₁ × ⁶C₂ + ³C₂ × ⁶C₁ + ³C₃ =3×15+3×6+1

= 45 + 18 + 1

= 64

We know that,

ⁿC_r

According to the question,

ⁿC_{r – 1} =36,

ⁿC_r =84,

ⁿC_{r +1}=126

2n-2r=3r+3

⇒2n – 3 = 5r … (i)

=3n-3r+3=7r

3n+3=10r … (ii)

From (i) and (ii),

We get,

2(2n – 3) = 3n+3

4n – 3n – 6 – 3=0

n=9

And r=3

Now

^rC₂=³C₂ = 3!/2!

=3

According to the question,

Digits that can be used = 3,5,7,8,9

Since, no digits can be repeated,

The number of integers is ⁵P₅ = 5! = 120

For a four-digit integer to be greater than 7000,

The four-digit integer should begin with 7,8 or 9.

The number of such integer = 3 × ⁴P₃ =3 × ⁴P₃ = 3(24) =72

Therefore, the total no of ways = 120+72 = 192

Let the number of intersection point of first line =0

Let the number of intersection point of 2^nd line = 1

Let the number of intersection point of 3^rd line = 2+1

Let the number of intersection point of 4^th line = 3+2+1

. . .

Let the number of intersection point of n^th line =(n-1) +(n-2)…..(3)(2)(1), where n=20

S = ( n – 1) × n/2

=19×10

=190

According to the question,

All telephone numbers have six digits

Given that,

The first two digits = 41 or 42 or 46 or 62 or 64

Hence, the number of 2 digits that the telephone number begins with =5

First, two digits can be filled in 5 ways,

The remaining four-digits can be filled in ⁸P₄ ways,

⁸P₄ = 8!/(8 – 4)! = 1680

Therefore, the number of telephone numbers having six distinct digits =5×1680

=8400

We know that,

ⁿC_r

According to the question,

Total number of questions =5

Number of questions to be answered =4

Compulsory questions are question number 1 and 2

Hence, the number of ways in which the student can make the choice = ³C₂

³C₂ = 3!/(2!1!) =3 ways

We know that,

ⁿC_r

Let the number of sides the given polygon have = n

Now,

The number of line segments obtained by joining n vertices = ⁿC₂

So, number of diagonals of the polygon = ⁿC₂ – n = 44

n² – 3n – 88 = 0

(n – 11) (n + 8) = 0

n = 11 or n = – 8

The polygon has 11 sides.

Long Answer Type