Inclusion and Exclusion Principle in a Discrete Math Problems

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Suppose you have seven dice–each a different color of the rainbow; otherwise the dice are standard,with faces numbered 1 to 6. A roll is a sequence specifying a value for each die in rainbow(ROYGBIV) order. For example, one roll is(3,1,6,1,4,5,2)indicating that the red die showed a3, the orange die showed 1, the yellow 6, etc.

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Combinatorial Arguments Discrete Mathematics City College of San Francisco Outline • Counting Techniques: Peer Review Assignment Q&A • Combinatorial Arguments • Combinatorial Proofs • The Pigeonhole Principle Combinatorial Arguments Combinatorial Proofs • Rather than relying on algebraic techniques, you can sometimes prove the equality of two mathematical expressions by using each expression to count the same set. • Combinatorial Proofs: • • • • Define some set S. Determine |S| using some formulaic expression l. Determine |S| using some other formulaic expression r . Conclude l = r . A point of struggle is that S can present anything: strings, numbers, people, etc. Let S contain elements that are easy for you to count conceptually. zyBooks Exercise 11.2.4 (b) Prove the following proposition by using a combinatorial argument: Proposition For positive integers, n, m, and k , where k ≤ m ≤ n:  n m m k = n  n − k  k m−k . zyBooks Exercise 11.2.4 (b) - Sample Response Proof. We will argue the truth of this statement using a combinatorial argument. Consider the set A := {1, 2, 3, . . . , n}. This gives that |A| = n. Now, let’s count the number of ways to select two disjoint subsets B and C from A where |B| = k and |C| = m − k for positive integers, n, m, and k, where k ≤ m ≤ n. This means that |B ∪ C| = k + m − k = m. All such pairs of disjoint subsets B and C form the set S we want to count. Let’s count |S| in two ways.  n 1. There are m ways to pick the m elements for the two subsets, and there are m k ways to pick from those m elements the elements for B. There is no choice for the elements of C as they would be the remaining m − k elements selected. This n  m n  m means there are m ways to form these two subsets (i.e. |S| = m ). k k 2. Another approach is to count  the number of ways to pick the k elements for B from the n elements of A (i.e. kn ) and count the number of ways to select the m − k n−k  elements for C from the remaining n − k elements from A − B (i.e. m−k ). This  n−k   n−k  would give a total of kn m−k such subset formations (i.e. |S| = kn m−k ). This argument shows that  n m m k = n  n − k  k m−k .  The Pigeonhole Principle • The Pigeonhole Principle: If |A| > |B|, then for every function f : A → B defined on A, there exist two different elements of A that are mapped by f to the same element of B. • When you are trying to verify some statement using the Pigeonhole Principle, you want to identify • the set A, • the set B where |B| < |A|, and • the function f : A → B that is defined on all of A. • The Generalized Pigeonhole Principle: If |A| > k · |B|, then for every function f : A → B defined on A, there exist k + 1 different elements of A that are mapped by f to the same element of B. zyBooks Exercise 11.3.2 (a) There are 121.4 million people in the United States who earn an annual income that is at least $10,000 and less than $1,000,000. Annual income is rounded to the nearest dollar. Show that there are 123 people who earn the same annual income in dollars. zyBooks Exercise 11.3.2 (a) - Sample Response Proof. Let’s use the Generalized Pigeonhole Principle to resolve this statement. We’ll define the set A to represent the 121.4 million people. So, |A| = 121, 400, 000. Also, let’s define the set B to be the set of incomes. This means that |B| = 1000000 − 10000 = 99000. If we map A to B by assigning each person to their annual thenm the Pigeonhole Principle says that there will be at least l m income, l |A| 121400000 = = 123 people assigned to the same income as stated.  |B| 99000 WRITING PROMPT TWO 1. How many paths are there from point (0,0) to point (50,50) on the two-dimensional plane if each step along a path increments one coordinate and leaves the others unchanged and there is an impassable obstacle at the points (10,11) and (20,19)? Provide a formula representation of your result that includes binomial coefficients, and navigate the reader through the development of your formula. 2. Using the Pigeonhole Principle, prove that in every set of 100 integers, there exist two whose difference is a multiple of 37. Identify the function (including its domain and target) outlined in either of our class resources while explaining how the principle is being applied.
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Written Prompt 2
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Writing Prompt 2
1.
In determining the number of paths from point (0,0) to (50,50) in a two- dimensional plane, we
use the inclusion-exclusion principle. From point (0,0) to (50,50), there are (100
) paths in the
50
absence of obstructed paths. The number of steps from the point (0,0) to the final point is 100
steps obtained by adding 50 steps plus 50. We are given two obstacles at points (10,11) and
(20,19). From the first obstacle, we have (20
) paths from the starting point of the first obstacle.
10
There are (80
) from the first obstacle to the end obtained by subtracting (20
) from the total
40
10
number of paths from start to finish (100
). Thus, from the first obstacle, there are;
50
(20
) * (80
) paths.
10
40
From the second obstacle, the number of paths is obtained by adding up both the first and the
second obstacle and subtracting it from the total number of paths as shown below;
19
39
(20
) + (11
) = (21
)
10
39
(100
)- (21
)= (61
)
50
29
39
Thus, (21
) * (61
) paths go through the second obstacle.
29

The paths that are passing through the first and the second obstacle is given by
19
(20
) * (11
) * (61
)...


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