Number of Arrangements Without Element Repetition

Question 1. There 26 letters in the alphabet.

A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P, Q, R, S, T, U, V, W, X, Y, Z

We want to form a password of six characters. How many arrangements of six characters can we form? Repetition is not allowed.
(Repetition not allowed means none of the six characters can be repeated.)
Solution:
(Number of ways for 1st character)⋅(Number of ways for 2nd character)⋅( Number of ways for 3rd character)⋅(Number of ways for 4th character)⋅(Number of ways for 5th character)⋅(Number of ways for 6th character)
26×25×24×23×22×21
=165,765,600

Question 2. Given 5 flags of different colours, how many different signals can be generated if each signal requires the use of 2 flags, one below the other?
Solution 2-I:
Each signal requires the use of 2 flags.
There will be as many flags as there are ways of filling in 2 vacant places in succession by the given 5 flags of different colours.
The upper vacant place can be filled in 5 different ways by any one of the 5 flags following which, the lower vacant place can be filled in 4 different ways by any one of the remaining 4 different flags.
Thus, by multiplication principle, the number of different signals that can be generated is 5×4=20.
Solution 2-II:
There are 5 flags, number of flags to generate a signal equals 2. So, 1st flag can be selected by 5 ways, 2nd flag can be selected by 4 ways. None of a flag can be selected more than once into a signal.
Hence, the number of ways equals 5×4=20

🌈 Permutations, number of arrangements without elements repetition

Multiplication with Element Repetition — License Plates excluding Alphabet’s Vowels

Example 1: License Plates
Question: The license plate on a car consists of any 3 letters of the alphabet (excluding the vowels and followed by any 3 digits (0 to 9). For a car chosen at random, what is the probability that the license plate starts with a ’Y’ and ends with an odd digit?
Answer:
Step 1 : Identify what events are counted
The License plate starts with a ’Y’, so there is only 1 choice for the first letter, and ends with an even digit, so there are 5 choices for the last digit (1,3,5,7,9).
Step 2 : Find the number of events
Use the counting principle. For each of the other letters, there are 20 possible choices (26 in the alphabet, minus 5 vowels and ’Q’) and 10 possible choices for each of the other digits.
Number of events =1×20×20×10×10×5=200 000
Step 3 : Find the number of total possible license plates
Use the counting principle. This time, the first letter and last digit can be anything. Total number of choices =20×20×20×10×10×10=8 000 000
Step 4 : Calculate the probability
The probability is the number of events we are counting, divided by the total number of choices.

Example 2: License Plates
How many license plates can be made if each is to be three digits followed by 3 letters. The plate number cannot begin with a 0.
We can think of this as a sequence of tasks and apply the Fundamental Counting Principle:

🌈 Permutations with repetitions

Multiplication with Element Repetition — Codes

⛲ Example 1. A large school issues special coded identification cards that consist of two letters of the alphabet followed by three numerals. For example, AB 737 is such a code. How many different ID cards can be issued if the letters or numbers can be used more than once?
✍ Solution:
As the letters can be used more than once, then each letter position can be filled in 26 different ways, i.e. the letters can be filled in 26⋅26=676 ways. Each number position can be filled in 10 different ways; hence, the numerals can be filled in 10⋅10⋅10=1000 different ways. So, the code can be formed in 676⋅1000=676,000 different ways.

⛲ Ex2. A company uses a coding system to identify its clients. Each code is made up of two letters and a sequence of digits, for example AD108 or RR 45789.
The letters are chosen from A; D; R; S and U. Letters may be repeated in the code.
The digits 0 to 9 are used, but NO digit may be repeated in the code.
a. How many different clients can be identified with a coding system that is made up of TWO letters and TWO digits?
b. Determine the least number of digits that is required for a company to uniquely identify 700 000 clients using their coding system.
✍ Solution:
a. 5×5×10×9=2250
b.

No of digits used letters digits total
1 5×5 10 250
2 5×5 10×9 2250
3 5×5 10×9×8 18000
4 5×5 10×9×8×7 126000
5 5×5 10×9×8×7×6 756000

Codes of two letters and five digits will ensure unique numbers for 700 000 clients.

🌈 Permutation with replacement

Counting — Enumeration — Multiplication

Basic Principle of Enumeration (if a few elements) or Multiplication (if lots of elements)
1a. Addition Theorem:
If a work completes in a single level and we have different independent options to do it, then it can be completed by the number of ways equals to the addition theorem of basic principle of enumeration.
1b. Addition Principle
If first operation can be performed in m ways and another operation, which is independent of the first, can beperformed in n ways. Then, either of the two operations can be performed in m+n ways. This can be extended to any finite number of exclusife events.

2a. Multiplication theorem:
If a work completes in more then one level and we have different independent options to do it, then it can be completed by the number of ways equals to the product of the options called the multiplication theorem of basic principle of enumeration.
2b. Multiplication Principle
If first operation can be performed in m ways and then a second operation can be performed in n ways. Then, the two operations taken together can be performed in m×n ways. This can be extended to any finite number of operations.

Multiplication Principle:
Consider the 3 letter words that can be made from the letters WORD if no letter is repeated There are:
4 ways of choosing the 1st letter
3 ways of choosing the 2nd letter
2 ways of choosing the 3rd letter
Number of words =4×3×2=24

This is an illustration of the multiplication principle i.e., if several operations are carried out in a certain order, then the number of ways of performing all the operations is the product of the numbers of ways of performing each operation.

If on operation can be performed in m different ways and another operation in n different ways then these two operations can be performed one after the other in m×n ways.

Example 1. Baskin Robbins has 20 flavours of ice cream and 11 flavours of sherbet. In how many ways could you select
a. A scoop of ice cream or a scoop of sherbet?
b. A scoop of ice cream and then a scoop of sherbet?
🔑
a. 20+11=31
b. 20×11=220

Counting Sample Spaces
Below we will discuss how to find the number of sample points in a sample space or in an event.

Rule 1: Multiplication
If an experiment produces m results (outcomes) and for each result the second experiment produces n results then there is a total of m×n results.

Example 2
If two coins are tossed, then
the total number of sample points =2×2=4.
In general, if n, coins are tossed, the total sample points is 2n.

Example 3
If two dice are rolled, then
the total number of sample points =6×6=36.
In general, if n dice are rolled, the total sample points =6n.

Example 4: Enumeration
If a die is rolled and a coin is tossed simultaneously, then the total number of sample points =6×2=12, and the sample space is
S={1H, 2H, 3H, 4H, 5H, 6H, 1T, 2T, 3T, 4T, 5T, 6T}.
Note: H is a Head side and T is a Tail side.

Example 5
In a multiple choice test with 4 questions each of 3 choices, the number of ways a student can answer is 3×3×3×3=81.

🌈 Using tree diagrams to enumerate parallelable occurrence of an event

Rule 2: Permutations
Rule 3: Combinations
Rule 4: Venn Diagram

Counting — With Replacement — Without Replacement

Counting principles
Simple counting problems
This section will introduce you to some of the basic principles of counting. You will apply some of this in justifying the binomial theorem and in the chapter of Probabilities you will use these principles to tackle many probability problems. We will start with two examples.

Example 1
Nine paper chips each carrying the numerals 1-9 are placed in a box. Two chips are chosen such that the first chip is chosen, the number is recorded and the chip is put back in the box, then the second chip is drawn. The numbers on the chips are added. In how many ways can you get a sum of 8?
Solution:
To solve this problem, count the different number of ways that a total of 8 can be obtained:

1st chip 1 2 3 4 5 6 7
2nd chip 7 6 5 4 3 2 1

From this list, it is clear that you can have 7 different ways of receiving a sum of 8.

Example 2
Suppose now that the first chip is chosen, the number is recorded and the chip is not put back in the box, then the second chip is drawn. In how many ways can you get a sum of 8?
Solution:
To solve this problem too, count the different number of ways that a total of 8 can be obtained:

1st chip 1 2 3 5 6 7
2nd chip 7 6 5 3 2 1

From this list, it is clear that you can have 6 different ways of receiving a sum of 8.

The difference between the two situations is described by saying that the first random selection is done with replacement, while the second is without replacement, which ruled out the use of two 4s.

🌈 number of arrangements with element repetition or with no repetition

Enumeration

multiplication b

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