Hey, y'all!
Discrete math, am I right? It can be a real headache, but don't worry – I got you covered. I found some problems and questions that might help y'all out with this tricky subject.
Discrete Math Problems
Okay, this is one of those tricky discrete math problems that can really make your head spin. But don't worry, I'll break it down for y'all. Basically, we want to find a function f(x) such that:
f(0) = 1, f(1) = 2, f(2) = 4, and f(3) = 7.
So how do we go about doing that? Well, we can start by writing out the first few differences:
f(1) - f(0) = 1
f(2) - f(1) = 2
f(3) - f(2) = 3
Notice anything? The differences are increasing by 1 each time! So we can try a quadratic formula:
f(x) = ax^2 + bx + c
Now we just need to solve for a, b, and c. Using our initial conditions, we get:
a = 1/2, b = 3/2, c = 1
So our function is:
f(x) = (1/2)x^2 + (3/2)x + 1
Make-Up Exam Questions
Alright, now let's move on to some exam questions. Don't panic y'all – just take a deep breath and read the question carefully.
1. Let A = 1, 2, 3, 4 and B = 1, 2, 3, 4, 5. How many injective functions are there from A to B?
Remember that an injective function is a function where no two elements of the domain map to the same element of the range. The answer is 5!/(5-4)! = 120.
2. Let G be a simple graph with 6 vertices and 10 edges. What is the maximum number of vertices that can have degree 3?
This one's a bit trickier. We know that the sum of the degrees of all vertices in a simple graph is equal to twice the number of edges. So in this case, the sum of the degrees is 2*10 = 20. Let x be the number of vertices with degree 3. Then the sum of degrees of all vertices with degree greater than 3 is 6-x. So we have:
3x + (6-x) > 20
x > 7/2
So the maximum value of x is 3.
Hope these problems and questions help y'all out!
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