Chapter 7: Permutations and Combinations

We know that,

ⁿP_r

According to the question,

W₁ can occupy chairs marked 1 to 4 in 4 different way.

W₂ can occupy 3 chairs marked 1 to 4 in 3 different ways.

So, total no of ways in which women can occupy the chairs,

⁴P₂ = 4!/(4 – 2)!

= (4 × 3 × 2 × 1)/(2 × 1)

⁴P₂ = 12

Now, 6 chairs will be remaining.

M₁ can occupy any of the 6 chairs in 6 different ways,

M₂ can occupy any of the remaining 5 chairs in 5 different ways

M₃ can occupy any of the remaining 4 chairs in 4 different ways.

So, the total no of ways in which men can occupy the chairs,

⁶P₃ = 6!/(6 – 3)!

=120

Hence, total number of ways in which men and women can be seated

⁴P₂x⁶P₃ = 120 × 12

=1440

According to the question,

Arranging in alphabetical order, we get,

A C H I R T

Number of words that can start with A=5!

Number of words that can start with C=5!

Number of words that can start with H=5!

Number of words that can start with I=5!

Total = 5! +5!+5!+5!

= 120+120+120+120=480

Since, the word that can start with R is RACHIT

Number of words that can start with R = 1

Hence, we obtain that,

The rank of the word RACHIT = 480+1=481

We know that,

ⁿC_r

No of questions in group A=6

No of questions in group B=6

According to the question,

The different ways in which the questions can be attempted are,

Hence, the number of different ways of doing questions,

= (⁶C₂x⁶C₅)+(⁶C₃x⁶C₄)+(⁶C₄x⁶C₃)+(⁶C₅x⁶C₂)

= (15×6)+(20×15)+(15×20)+(6×15)

=780

We know that,

ⁿC_r

According to the question,

Number of points = 18

Number of Collinear points = 5

Number of lines form by 18 points = ¹⁸C₂

For 5 points to be collinear = ⁵C₂

The number of lines that can be formed joining the point, = ¹⁸C₂–⁵C₂+1

=153-10+1

=144

We know that,

ⁿC_r

According to the question,

Case 1:

If both A and B are selected =1x1x⁶C₄

Case 2:

If neither A nor B are selected = ⁶C₆ = 1

If B is selected but A is not selected = 1x⁶C₅

Adding the results of both A and B being selected, neither A nor B being selected and B being selected but A not being selected,

We get,

15+1+6=22

We know that,

ⁿC_r

Number of ways a chairperson can be selected =12

Selection of 4 other people =¹¹C₄

Selection of 5 people = 330 × 12

=3960

According to the question,

Number of letters in automobile license plates = 2

We know that,

There are 26 alphabets

So, letters can be arranged without repetition in the following number of ways,

= 26×25

=650

Number of digits in automobile license plates =3

We know that, there 10 digits

Hence the number of digits without repetitions

=10×9×8=720

Therefore, the total number of way automobile license plates

=720×650

=468000

We know that,

ⁿC_r

According to the question,

Number of black balls = 5

Number of red balls = 6

Number of ways in which 2 black balls can be selected = ⁵C₂

Number of ways in which 3 red balls can be selected =⁵C₃

Number of ways in which 2 black & 3 red balls can be selected

=⁵C₂×⁵C₃

=10×20

=200

Permutations of n distinct things taken r together =ⁿC_r

And when 3 particular things must occur together, we get,

= ^{n – 3}C_{r – 3}

= ^{n – 3}C_{r – 3} × (r – 2)! × 3!

We know that,

ⁿP_r

According to the question,

Total number of vowels letter =3,

Total number of consonants letter =5

The vowels can be placed in

⁶P₃ = 6!/3! = 120

The number of ways consonants can be arranged placed =5! =120

Therefore, total number of ways it can be arranged =5!×⁶P₃=120×120 =14400