Basic fundamental type questions of permutation

1.     In a test paper first section contains 6 questions each with 5 choices and 2^nd section contains 5 questions each with 4 choices. In how many different ways can the paper be answered if all the questions are attempted?

Solution:

1^st section = 5x5x5x5x5x5 = 5^6 ways

2^nd section= 4x4x4x4x4 = 4^5

Therefore the whole paper can be answered = 5^6x4^5

2.     5 persons entered the lift cabin on the ground floor of a 7 floor office. Suppose each of them can leave the cabin independently at any floor beginning with the 1^st. Find the total number of ways in which each of the 5 persons can leave the cabin at different floors.

Solution:

A1 A2 A3 A4 and A5 are the 5 persons.

A1 can leave at any of the 6 floor. A2 can leave at remaining any of the 5. Similarly A3 can leave at any of the 4 floor, A4 can leave at any of the 3 floor and A5 leave at any of the 2 floor.

Thus the total number of ways = 6x5x4x3x2 = 720

Circular Permutations

In circular permutations the things are arranged in a circle or ring form.

Hence the number of circular permutations of an n objects = n!/n = (n-1)!

Problems

1.     In how many ways can 5 boys form a ring?

Solution:

            5 boys form a ring = (n-1)!  =  (5-1)! = 4! = 24

2.     In how many ways can 6 men and 3 ladies be arranged at a round table if 3 ladies are never together?

Solution:

Total number of arrangement = 6men + 1 (3 ladies considered as 1) = 7! = 5040

Number of arrangement of 3 ladies together with 6 men = 3x6! = 2160

Therefore number of arrangements in which 3ladies are never together

= 5040-2160 = 2880