Home > Practice > Arithmetic Aptitude > Permutations and Combinations > Miscellaneous
31. A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?
In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.
So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 24 - 1
Similarly for the 2nd section there are 25 - 1 ways in which he can attempt and for the 3rd section there are 26 - 1 ways.
The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.
Thus, total number of ways in which he can attempt questions in that paper:
= (24 - 1)(25 - 1)(26 - 1)
= 15 × 31 × 63
= 29295
Solution:
At least 1 question from each section is compulsory, so from the 1st section the candidate can attempt 1 or 2 or 3 or 4 questions.In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it.
So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 24 - 1
Similarly for the 2nd section there are 25 - 1 ways in which he can attempt and for the 3rd section there are 26 - 1 ways.
The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections.
Thus, total number of ways in which he can attempt questions in that paper:
= (24 - 1)(25 - 1)(26 - 1)
= 15 × 31 × 63
= 29295
32. In how many ways a President, VP and Water-boy can be selected from a group of 10 people.
Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be 10P3
Solution:
We are selecting three different posts here, so order matters.Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be 10P3
33. In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:
= nC2 = 153
On solving we get, n = −17 and n = 18
It cannot be negative so n = 18
Solution:
Let there were x teams participating in the games, then total number of matches,= nC2 = 153
On solving we get, n = −17 and n = 18
It cannot be negative so n = 18
34. A box contains 10 balls out of which 3 are red and 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?
One red ball and 5 blue balls
OrTwo red balls and 4 blue balls.
Total number of ways,
= (3C1 × 7C5) + (3C2 × 4C7)
= 63 + 105
= 168
Solution:
Six balls can be selected in the following ways,One red ball and 5 blue balls
OrTwo red balls and 4 blue balls.
Total number of ways,
= (3C1 × 7C5) + (3C2 × 4C7)
= 63 + 105
= 168
35. Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
= 3C2 × 4! × 4!
= 1728
Solution:
Required number of ways,= 3C2 × 4! × 4!
= 1728
36. A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,6), covering the shortest possible distance.
The number of ways in which these steps can be taken is given by,
Solution:
In order to reach (5,6) covering the shortest distance at the same time the man has to make 5 horizontal and 6 vertical steps.The number of ways in which these steps can be taken is given by,
37. 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?
Solution:
Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.
38. In an examination paper, there are two groups each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?
2 question from first group and 3 question from second groupOr3 question from first group and 2 question from second group.
= (4C2 × 4C3) + (3C4 × 4C2)
= 24 + 24
= 48
Solution:
5 questions can be selected in the following ways,2 question from first group and 3 question from second groupOr3 question from first group and 2 question from second group.
= (4C2 × 4C3) + (3C4 × 4C2)
= 24 + 24
= 48
39. After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?
Then,
Solution:
Let total number of persons present in the party be x,Then,
40. How many diagonals can be drawn in a pentagon?
Total number of sides and diagonals,
= 5C2
=
= 5 × 2 = 10
This includes its 5 sides also.
⇒ Diagonals = 10 – 5 = 5
Hence, the number of diagonals
= 10 – 5
= 5
Solution:
A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs.Total number of sides and diagonals,
= 5C2
=
= 5 × 2 = 10
This includes its 5 sides also.
⇒ Diagonals = 10 – 5 = 5
Hence, the number of diagonals
= 10 – 5
= 5