Permutation-Combination Exercise 3

A combination of four vertices is equivalent to one interior point of intersection of diagonals.

∴ No. of interior points of intersection = nC₄ = 70

⇒ n (n – 1) (n – 2) (n – 3)

= 5. 6. 7. 8

∴ n = 8 So, number of diagonals

= ⁸C₂ – 8 = 20

Let the Vice-chairman & the Chairman form 1 unit.

Along with the eight directors, we now have to arrange 9 different units in a circle.

This can be done in 8! ways.

At the some time, the Vice-chairman & the chairman can be arranged in two different ways.

Therefore, the total number of ways = 2 × 8!

The no. of ways are

By option elimination,

2n + 1 = 7. So n = 3.

Consider the number:

x y z where x < y , z < y and x ≠ 0.

If y = 9 , x can be between 1 to 8 and z can be between 0 to 8

Total combinations = 9 × 8 = 72

If y = 8 , x can be between 1 to 7 and z can be between 0 to 7

combinations = 7 × 8 = 56

Similarly we add all combinations :

8 × 9 + 7 × 8 + 6 × 7 + 5 × 6 + 4 × 5 + 3 × 4 + 2 × 3 + 2 × 1

= 240 ways.

As they are consecutively numbered total number of ways will be

6 + 5 + 4 + 3 + 2 + 1 = 21 ways

Let number of girls = x and the number of boys = y

45 games in which both the players were girls

∴ x = 10

190 games, where both the players were boys.

yC₂ = 190

⇒ y(y – 1) = 380

∴ y = 20

Hence the total number of games in which one player was a boy and the other was a girl

= 10 × 20 = 200

For each person to know all the secrets the communication has to be between the Englishmen (who knows say E1 French) and one Frenchmen (say F1).

The other two in each case will communicate with E1 & F1 respectively.

So for minimum no. of calls, E2 gives information to E1 & receives it after E1 interacts with F1.

So 2 calls for each of the four E2, E3, F2 and F3,

i.e., 8 calls +1 call (between E1 & F1). Hence 9 calls in all.

Number of ways of selecting 5 guests from nine friends = ⁹C₅

Out of these, ⁷C₃ ways are those in which two of the friends occur together [3 more persons to be selected out of remaining 7]

∴ Number of ways, in which two of the friends will not attend the party together

= ⁹C₅ – ⁷C₃ = 91.

Let there be n participants in the beginning.

Then the number of games played by (n – 2) players

= n – ²C₂

(Two players played three games each)

⇒ (n - 2)(n - 3) = 156

⇒ n² - 5n - 150 = 0

⇒ n = 15

The required number of ways

=

=

=

............. = .

Let the sides of the game be A and B. Given 5 married couples, i.e., 5 husbands and 5 wives.

Now, 2 husbands for two sides A and B can be selected out of 5 = ⁵C₂ = 10 ways.

After choosing the two husbands their wives are to be excluded (since no husband and wife play in the same game).

So we are to choose 2 wives out of remaining 5 – 2 = 3 wives

i.e., ³C₂ = 3 ways.

Again two wives can interchange their sides A and B in 2! = 2 ways.

By the principle of multiplication, the required number of ways

= 10 × 3 × 2 = 60

Six students S₁, S₂, ........, S₆ can be arranged round a circular table in 5 ! ways.

Among these 6 students there are six vacant places, shown by dots (•) in which six teachers can sit in 6 ! ways.

Hence, number of arrangement = 5 ! × 6 !

Number of ways of arranging P big animals into

m – n big cages = ^m–nP_p.

Now remaining animals can be arranged in any cage in

^m–pP_m–P ways

∴ Desired number of ways

= ^m–nP_p × ^m–pP_m–p

Two possibilities are there :

(i) Chemistry part I is available in 8 books with Chemistry part II.

or

(ii) Chemistry part II is available in 8 books but Chemistry part I is not available.

Total No. of ways

= 1 × ⁶C₁ + ⁷C₃

=

As all the points are equally spaced, the area of all the convex pentagons will be the same.

No. of words which have at least one letter repeated = total no. of words – total no. of words in which no letter is repeated

= 10⁵ – ¹⁰P₅

= 100000 – 10 × 9 × 8 × 7 × 6

= 100000 – 30240 = 69760

Under the given restrictions, 5 questions can be selected in the following ways :

2 questions from the first section and 3 questions from the second section

Or

3 questions from the first section and 2 questions from the second section.

Required no. of ways

= ⁴C₂ × ⁴C₃ + ⁴C₃ × ⁴C₂

= 24 + 24 = 48

Six balls can be selected in the following ways: one red balls and 5 blue balls or

Two red balls and 4 blue balls

Total number of ways

= ³C₁ × ⁷C₅ + ³C₂ × ⁷C₄

= 63 + 105 = 168

Two tallest boys can be arranged in 2! ways. Rest 18 can be arranged in 18! ways.

Girls can be arranged in 6! ways.

Total number of ways of arrangement

= 2! × 18! × 6!

= 18! × 2 × 720 = 18! × 1440

To construct 2 roads, three towns can be selected out of 4 in 4 ×3×2 = 24 ways.

Now if the third road goes from the third town to the first town, a triangle is formed, and if it goes to the fourth town, a triangle is not formed. So there are 24 ways to form a triangle and 24 ways of avoiding a triangle.