DISCRETIVE MATHAMATICS
1Q:- How many different bit strings of length seven are there?
Solution:
Each of the seven bits can be chosen in two ways, because each bit is either 0 or 1. Therefore, the product rule shows there are a total of 27 = 128 different bit strings of length seven.
2Q:-β’ Example:
How many different license plates can be made if each plate contains a sequence of three uppercase English letters followed by three digits?
β’ Solution:
STEP:-1 THREE LETTERS AND THREE DIGITS SO TOTAL 6 PLACES__ __ __ __ __ __
STEP:-2 first place have 26 choices any one letter can write it place
STEP:-3 second place also same
STEP:-5 4th place will filled with digits have 10 choices can fill any one of them
26x26x26x10x10x10 = 17,576,000
17,576,000 possible license plates
There are 26 choices for each of the three uppercase English letters and ten choices for each of the three digits. Hence, by the product rule there are a total of
26 Β· 26 Β· 26 Β· 10 Β· 10 Β· 10 = 17,576,000 possible license plates.
3Qβ’ Example:
Suppose that either a member of the mathematics faculty or a student who is a mathematics major is chosen as a representative to a university committee. How many different choices are there for this representative if there are 37 members of the mathematics faculty and 83 mathematics majors and no one is both a faculty member and a student?
β’ Solution:
STEP:-1 can choose representative of committe one person of student either faculty any one of them
STEP:-2 there are 37 members of faculty and 83 persons of students
STEP:-3 how many diffirent choices to select the committee persons is total is 37+83 choices is there
STEP:-4 pick one person only so the answer is 120 choices to pick or select any of them
There are 37 ways to choose a member of the mathematics faculty and there are 83 ways to choose a student who is a mathematics major. Choosing a member of the mathematics faculty is never the same as choosing a student who is a mathematics major because no one is both a faculty member and a student. By the sum rule it follows that there are 37 + 83 = 120
possible ways to pick this representative.
5QExample:
β’ Solution:
WE HAVE USED THIS FORMULA P(n,r)=n(n β 1)(n β 2) Β· Β· Β· (n β r + 1)
STEP: 1 REMEBER THAT ABOVE FORMULA
STEP :2 NUMBER PEOPLES PARTCIPATE IN CONTEST (n)=100
STEP :3 THRE ARE THREE PRIZES FROM 100 PEOPLES SO
STEP :4 FIRST PRIZE GIVEN BY ONE PERSON OUT OF 100
SECOND PRIZE GIVEN BY ONE PERSON OUT OF 99
P(100, 3) = 100 Β· 99 Β· 98 = 970,200.
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24/06/2021 ZOOM CLASS
6Qβ’ Example:
Suppose that a saleswoman has to visit eight different cities. She must begin her trip in a specified city, but she can visit the other seven cities in any order she wishes. How many possible orders can the saleswoman use when visiting these cities?
β’ Solution:
STEP :-1 THE WOMEN HAS TO VISIT EIGHT CITIES BUT SHE MUST START ANY ONE OF THESE CITIES
SO SHE VISIT 7 CITIES
STEP :-2 THE PLACES ARE 7 THE POSSIBLE WAYS ARE 7!
STEP:-3 __ __ __ __ __ __ __
7 x 6 x 5 x 4 x 3 x 2 x 1
STEP :- 4 7 x 6 x 5 x 4 x 3 x 2 x 1 = 5040
{ OR }
The number of possible paths between the cities is the number of permutations of seven elements, because the first city is determined, but the remaining seven can be ordered
arbitrarily. Consequently, there are
7! = 7 Β· 6 Β· 5 Β· 4 Β· 3 Β· 2 Β· 1 = 5040
ways for the sales woman to choose her tour. If, for instance, the saleswoman wishes to find the path between the cities with minimum distance, and she computes the total distance for each possible path, she must consider a total of 5040 paths!7Qβ’ Example:
How many permutations of the letters ABCDEFGH contain the string ABC ?
β’ Solution:
STEP :-1 THE GIVEN LETTERS ARE ABCDEFGH TOTAL 8 LETTERS CONTAINS A STRING ABC
STEP :- 2 IN GIVEN LETTERS ALREADY A STRING ABC CAN MUST FORM A BLOCK
STEP :- 3 SO REMAINING D, E, F, G, H ARE 5 LETTERS TOTAL HAVE 6 BLOCKS
THE POSSIBLE WAYS ARE 6!
STEP:- 5 ___ __ __ __ __ __
6 x 5 x 4 x 3 x 2 x 1
STEP:- 6 6 x 5 x 4 x 3 x 2 x 1 = 720
{ OR }
Because the letters ABC must occur as a block, we can find the answer by finding the number of permutations of six objects, namely, the block ABC and the individual letters D, E,F, G, and H. Because these six objects can occur in any order, there are
6! = 720
permutations of the letters ABCDEFGH in which ABC occurs as a block.
5 x 4 x 3 x 2 x 1
4 x 3 x 2 x 1
5 x 4 x 3 x 2 x 1
4 x 3 x 2 x 1
3 x 2 x 1
26/06/2021 ZOOM CLASS
Combinations
a combination is a selection of items from a collection, such that (unlike permutations) the order of selection does not matter.
THEOREM:
How many ways are there to select five players from a 10-member tennis team to make a trip to a match at another school?
β’ Solution:
The answer is given by the number of 5- combinations of a set with 10 elements.
β’ Solution:
The number of ways to select a crew of six from the pool of 30 people is the number of 6-combinations of a set with 30 elements, because the order in which these people are chosen does not matterβ’ Solution:
How many bit strings of length eight either start with a 1 bit or end with the two bits 00?
β’ Solution:
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