CSC102 - DISCRETE STRUCTURES
DUE DATE: 03-05-2021
TOTAL MARKS: 15
Instructions:

Assignment should be hand written or typed on A4 page size, with front page having the following
details. (Note: Failing to attach the front page with the following details will result in deduction of
5 marks)
Reg. # :_____________
Name : _____________
Course Title : _________
Section : __________
Assignment # : _______
Submitted to : _________
Date : ___________
(Font size 16, Times New Roman)
 Questions should solve in order as given in assignment.
 No marks for late submission.
 Assignment should be well formatted.
 While solving each question, do show all intermediate steps.
Question 1:
[2]
A. Suppose that there are 7 air routes to travel from Turkey to Italy and 5 from Italy to
Malaysia.
i.
How many ways are there to plan a trip route for round-trip form Turkey to Malaysia via
Italy?
Ans. From turkey to Italy= 7 ways
From Italy to Malaysia= 5 ways
From Malaysia to Italy= 5 ways
From Italy to Turkey= 7 ways
So total ways of round trip= 7*5*5*7
Total ways=1225 ways
ii.
How many ways are there to plan a trip route for round-trip form Turkey to Malaysia via
Italy, without using a route more than once?
Ans. From Turkey to Italy= 7 ways
From Italy to Malaysia= 5 ways
As routes are used once so,
From Malaysia to Italy= 4 ways
From Italy to Turkey= 6 ways
Using Multiplication rule:
Total ways= 7*5*4*6
Total ways=840 ways
B. How many strings of six letters are there, using the English letters?
If starting and ending letters are same and letters can’t be repeated?
i.
Ans. As first and last letters are same so possibility= 26
For second letter 25 possibilities
For third letter= 24 possibilities
For fourth letter 23 possibilities
For fifth letter = 22 possibilities
By multiplication rule:
Total Strings= 26*25*24*23*22
Total String= 7893600
ii.
If it starts with O and letters can be repeated?
Ans. As first letter start with o so it has only 1 possibility.
As all other lteers can repeat so,
By Multiplication rule:
Total Strings= 26*26*26*26*26
Total String= 11881376
iii.
If the letter X is at each even position and no letters can be repeated?
Ans. Letter x is at position 2,4,6 so,
For first letter = 25 possibilities
For third letter = 24 possibilities
For fifth letter= 23 possibilities
By Multiplication rule:
Total Possibilities= 25*1*24*1*23*1
Total possibilities= 13800
[4x2=8]
iv.
If either starts with letter X or end with letter Y and the letters can be repeated?
Ans. If it start with X: 1 26 26 26 26 26
If it ends with Y: 26 26 26 26 26 1
If it starts and end with X and Y: 1 26 26 26 26 1
By inclusion exclusion Principle:
Total possibilities= (26)5 + (26)5-(26)4
Total possibilities= 23762752 – 456976
Total possibilities= 23305776
Question 2:
Ans. Basis Step:
As Z+>0 so,
P(1)=1+2
P(1)=3
P(1) is true when it is divisible by 3.
Induction step:
P(k)= k3 + 2k
P(k+1)= (k+1)3+2(k+1)
=k+3k+3k+1+2k+2
=k+3k+5k+3
=