Virtual University Solved Assignments Online Quiz Past Papers GDB VUsolutions VU solutions

Question 1; Mark: 10

Use Mathematical Induction to prove that


Solution:


Induction given statement is true for all set of whole numbers.




Question 2; Mark: 5

If is odd then is even (prove by contradiction).



Sol: Suppose 3x + 1 is odd and x is not even

x is not even à x is odd (x is either even or odd but not both)

à x = 2k + 1 ( x is not divisible by 2)

à 3x + 1 = 3 ( 2k +1) + 1 (substitute 2k +1 from x)

3x +1 = 6k + 3 + 1

= 6k + 4

= 2 ( 3k + 2) (Divisible by 2)

à 3x + 1 is even which is contradiction

Therefore 3x + 1 is odd then x is even

Question 3; Marks: 5

Find the number of arrangements of balls having colors Red, Yellow, Blue Green and Pink such that Red and Green balls must place next to each other.

Sol:

Red, Yellow, Blue, Green, Pink
Total numbers of balls = 5
Considering two balls Red and Green as a single ball ( due to restriction), we have to make the arrangements of 4 balls taken all together
No. of Arrangements = 4!
= 4 × 3 × 2 × 1
= 24
But two balls (restricted) permuted among themselves in two ways
i.e. 2! = 2 × 1 = 2 ways
Therefore total numbers of arrangements = 4! × 2!
= 24 × 2

= 48