This document provides a summary of topics related to counting, mathematical induction, and discrete probability. It covers basics of counting using rules like product and sum. It also discusses pigeonhole principle, permutations, combinations, and inclusion-exclusion principle. Additionally, it explains mathematical induction and concepts in probability like sample space, events, conditional probability, and Bayes' theorem. Examples are provided for various counting techniques and probability calculations.

1. COUNTING, MATHEMATICAL
INDUCTION AND DISCRETE
PROBABILITY
BY: SURBHI SAROHA

2. SYLLABUS
• Basics of Counting
• Pigeonhole Principal
• Permutations and Combinations
• Inclusion-Exclusion Principle
• Mathematical Induction
• Probability
• Bayes’ Theorem

3. BASICS OF COUNTING
• Assume we have a set of objects with certain properties.
• • Counting is used to determine the number of these objects.
• Examples:
• • Number of available phone numbers with 7 digits in the local calling area
• • Number of possible match starters (football, basketball) given the number of team members and their
positions
• Approach: – simplify the solution by decomposing the problem
• • Two basic decomposition rules:
• – Product rule • A count decomposes into a sequence of dependent counts (“each element in the first
count is associated with all elements of the second count”)
• – Sum rule • A count decomposes into a set of independent counts (“elements of counts are
alternatives”)

4. PIGEONHOLE PRINCIPAL
• Assume you have a set of objects and a set of bins used to store objects.
• • The pigeonhole principle states that if there are more objects than bins then there is at
least one bin with more than one object.
• • Example: 7 balls and 5 bins to store them
• • At least one bin with more than 1 ball exists.

5. EXAMPLE:
• • How many students, each of whom comes from one of the 50 states, must
be enrolled in a university to guaranteed that there are at least 100 who come
from the same state?
• Answer:
• • Let there by 50 boxes, one per state.
• • We want to find the minimal N so that N/50=100.
• • Letting N=5000 is too much, since the remainder is 0.
• • We want a remainder of 1 so that let N=50*99+1=4951

6. GENERALIZED PIGEONHOLE PRINCIPLE THEOREM
• If N objects are placed into k bins then there is at least one bin containing at
least N / k objects.
• Example.
• Assume 100 people. Can you tell something about the number of people
born in the same month. • Yes. There exists a month in which at least
• 100 / 12 = 8.3 = 9 people were born.

7. PERMUTATIONS
• A permutation of a set of distinct objects is an ordered arrangement of the
objects.
• Since the objects are distinct, they cannot be selected more than once.
• Furthermore, the order of the arrangement matters.
• Example:
• • Assume we have a set S with n elements.
• S={a,b,c}.
• • Permutations of S:
• • a b c a c b b a c b c a c a b c b a

8. EXAMPLE 1
• • How many permutations of letters {a,b,c} are there?
• • Number of permutations is:
• P(n,n) =P(3,3) =3! = 6
• • Verify: abc acb bac bca cab cba

9. EXAMPLE 2
• • How many permutations of letters A B C D E F G H
contain a substring ABC.
• Idea: consider ABC as one element and D,E,F,G,H as other
5 elements for the total of 6 elements.
• Then we need to count the number of permutation of
these elements.
• 6! = 720

10. EXAMPLE:
• Suppose that there are eight runners in a race. The winner
receives a gold medal, the second-place finisher receives a silver
medal, and the third-place finisher receives a bronze medal. How
many different ways are there to award these medals, if all
possible outcomes of the race can occur?
• Answer:
• note that the runners are distinct and that the medals are ordered.
• The solution is P(8,3) = 8 * 7 * 6 = 8! / (8-3)! = 336.

11. COMBINATIONS
• A k-combination of elements of a set is an unordered selection of k elements
from the set.
• Thus, a k-combination is simply a subset of the set with k elements.
• Example: • 2-combinations of the set {a,b,c}

12. COMBINATIONS
Example:
• We need to create a team of 5 player for the competion out of 10 team members. How
many different teams is it possible to create?
Answer:
• When creating a team we do not care about the order in which players were picked.
It is is important that the player is in.
Because of that we need to consider unordered sets of combinations.
• C(10,5) = 10!/(10-5)!5! = (10.9.8.7.6) / (5 4 3 2 1) = 2.3.2.7.3= 6.14.3= 6.42= 252

13. INCLUSION-EXCLUSION PRINCIPLE
• Used in counts where the decomposition yields two count tasks with overlapping elements.
• • If we used the sum rule some elements would be counted twice Inclusion-exclusion principle: uses a
sum rule and then corrects for the overlapping elements.
• We used the principle for the cardinality of the set union.

14. MATHEMATICAL INDUCTION
• Mathematical Induction is a mathematical technique which is used to prove a
statement, a formula or a theorem is true for every natural number.
• The technique involves two steps to prove a statement, as stated below −
• Step 1(Base step) − It proves that a statement is true for the initial value.
• Step 2(Inductive step) − It proves that if the statement is true for the
nth iteration (or number n), then it is also true for (n+1)th iteration ( or
number n+1).

15. PROBABILITY
• Probability can be conceptualized as finding the chance of occurrence of an
event.
• Mathematically, it is the study of random processes and their outcomes.
• The laws of probability have a wide applicability in a variety of fields like
genetics, weather forecasting, opinion polls, stock markets etc.

16. STEPS TO FIND THE PROBABILITY
• Step 1 − Calculate all possible outcomes of the
experiment.
• Step 2 − Calculate the number of favorable outcomes of
the experiment.
• Step 3 − Apply the corresponding probability formula.

17. THROWING A DICE
• When a dice is thrown, six possible outcomes can be on the top − 1,2,3,4,5,6.
• The probability of any one of the numbers is 1/6
• The probability of getting even numbers is 3/6 = 1/2
• The probability of getting odd numbers is 3/6 = 1/2

18. CONDITIONAL PROBABILITY
• The conditional probability of an event B is the probability that the event will
occur given an event A has already occurred. This is written as P(B|A).
• Mathematically − P(B|A)=P(A∩B)/P(A)
• If event A and B are mutually exclusive, then the conditional probability of
event B after the event A will be the probability of event B that is P(B).

19. PROBLEM 1
• In a country 50% of all teenagers own a cycle and 30% of all teenagers own a bike and
cycle. What is the probability that a teenager owns bike given that the teenager owns a
cycle?
• Solution
• Let us assume A is the event of teenagers owning only a cycle and B is the event of
teenagers owning only a bike.
• So, P(A)=50/100=0.5 and P(A∩B)=30/100=0.3 from the given problem.
• P(B|A)=P(A∩B)/P(A)=0.3/0.5=0.6
• Hence, the probability that a teenager owns bike given that the teenager owns a cycle is
60%.

20. BAYES’ THEOREM
• Bayes' theorem is stated mathematically as the following equation

21. THANK YOU