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Permutation-Combination Exercise 3

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Permutation and Combination 3

  • This online quiz will test your knowledge of Permutation and Combination in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

In a polygon no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon be 70 then the number of diagonals of the polygon is

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

permutation-and-combination-q-53577.png

∴ 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

In how many ways can the eight directors, the Vice-chairman and the Chairman of a firm be seated at a round-table, if the Chairman has to sit between the Vice-chairman and the Director?

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!

Out of 2n+1 students, n students have to be given the scholarships. The number of ways in which at least one student can be given the scholarship is 63. What is the number of students receiving the scholarship?

The no. of ways are

permutation-and-combination-q-53860.png

By option elimination,

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

How many three digit positive integers, with digits x, y and z in the hundred’s, ten’s and unit’s place respectively, exist such that x < y, z < y and x ≠ 0?

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.

There are 6 boxes numbered 1,2,.........6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is

As they are consecutively numbered total number of ways will be

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

In a chess competition involving some boys and girls of a school, every student had to play exactly one game with every other student. It was found that in 45 games both the players were girls and in 190 games both were boys. The number of games in which one player was a boy and the other was a girl is

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

45 games in which both the players were girls

permutation-and-combination-q-53841.png

permutation-and-combination-q-53835.png

∴ 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

Three Englishmen and three Frenchmen work for the same company. Each of them knows a secret not known to others. They need to exchange these secrets over person-to-person phone calls so that eventually each person knows all six secrets. None of the Frenchmen knows English, and only one Englishman knows French. What is the minimum number of phone calls needed for the above purpose?

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.

A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together is

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.

In a chess tournament, where the participants were to play one game with another, two chess players fell ill, having played 3 games each. If the total number of games played is 84, the number of participants at the beginning was

Let there be n participants in the beginning.

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

= n – ²C₂

permutation-and-combination-q-54067.png

(Two players played three games each)

permutation-and-combination-q-54061.png

⇒ (n - 2)(n - 3) = 156

⇒ n² - 5n - 150 = 0

⇒ n = 15

There are 10 bags B₁, B₂, B₃, ...., B₁₀, which contain 21, 22, ...., 30 different articles respectively. The total number of ways to bring out 10 articles from a bag is

The required number of ways

= permutation-and-combination-q-54055.png

= permutation-and-combination-q-54049.png permutation-and-combination-q-54042.png

= permutation-and-combination-q-54036.png

permutation-and-combination-q-54030.png

............. = permutation-and-combination-q-54024.png.

The number of ways in which a mixed doubles game in tennis can be arranged from 5 married couples, if no husband and wife play in the same game, is

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 teachers and six students have to sit round a circular table such that there is a teacher between any two students. The number of ways in which they can sit is

permutation-and-combination-q-54018.png

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 !

m distinct animals of a circus have to be placed in m cages, one in each cage. If n (< m) cages are too small to accommodate p (n < p < m) animals, then the number of ways of putting the animals into cages are

Number of ways of arranging P big animals into

m – n big cages = m–nPp.

Now remaining animals can be arranged in any cage in

m–pPm–P ways

∴ Desired number of ways

= m–nPp × m–pPm–p

A boy has 3 library cards and 8 books of his interest in the library. Of these 8, he does not want to borrow chemistry part II unless Chemistry part I is also borrowed. In how many ways can he choose the three books to be borrowed?

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

= permutation-and-combination-q-54012.png

There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?

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

Ten different letters of an alphabet are given. Words with 5 letters are formed from these given letters. Then the number of words which have at least one letter repeated is :

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

In an examination paper there are two sections each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any particular section. In how many ways can 5 questions be selected?

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

A box contains 10 balls out of which 3 are red and the rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour?

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

permutation-and-combination-q-54006.png

= 63 + 105 = 168

A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?

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

A,B,C and D are four towns any three of which are non-collinear. Then the number of ways to construct three roads each joining a pair of towns so that the roads do not form a triangle is

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.

Now check your Result..

Your score is

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