University:
Stanford University
Course:
MATH 20 | Calculus
Academic year:

2023
Views:

17

Pages:

2
Author:

MathPaladin

Questions and Answers Sheet 1 Discrete math Question #1 Is 2 in {2} ? Answer: In the set-rooster notation, all the elements of a set exist between the braces. As 2 is existed in between the braces of a set {2}, then 2 is an element in the set {2}. Hence, answer of the question ''Is 2 in {2} ?'' : Yes. Question #2 The number of terms in (𝑥 + 𝑦 + 𝑧)30 𝑖𝑠 496. How many terms would there be if the same terms were combined? How can we find out? Answer: Step 1 First note that every 𝑥 𝑎 𝑦 𝑏 𝑧 𝑐 with 𝑎 + 𝑏 + 𝑐 = 30 appears. One way of seeing this is observing  30 ( x + y + z )30 at (0, 0, 0) is 30! , which is not zero. So for a given value of a, every possible (𝑏, 𝑐)  ax  by  cz with 𝑏 + 𝑐 = 30 − 𝑎 can happen. Since b goes from 0 to 30 − 𝑎 , there are 31 − 𝑎 possibilities for b, each of which forces c to have the single value 30 − 𝑎 − 𝑏 Thus for a given a there are 31 − 𝑎 different possible (𝑏, 𝑐). Adding this over all a from 0 to 31 this gives the sum of 31 + 30 + ⋯ + 0 = 496 Question #3 If x, y, and z are natural numbers, how many solutions are there to 𝑥 + 𝑦 + 𝑧 = 25 ? Answer: Step 1 Here is a simpler explanation: Lets look at the equation 𝑥 + 𝑦 = 𝑘 where 𝑥 ≥0&𝑦 ≥0 the solution set is of form {{𝑘, 0}, {𝑘 − 1, 1}, ⋯ {0, 𝑘 + 1}} and there are (𝑘 + 1) items in the set. 𝑥 + 𝑦 + 𝑧 = 𝑘 can be broken down into: 𝑥 + (𝑦 + 𝑧) = 𝑘 If we choose 𝑥 = 0 , solutions of (𝑦 + 𝑧) = 𝑘 are 𝑘 + 1 . If we choose 𝑥 = 1 , solutions of (𝑦 + 𝑧) = 𝑘 − 1 are k. If we choose 𝑥 = 2 , solutions of (𝑦 + 𝑧) = 𝑘 − 2 are 𝑘 − 1 . If we choose 𝑥 = 𝑘 , solutions of (𝑦 + 𝑧) = 0 are 1. This is a simple natural number sum: 1 + 2 + 3 + ⋯ + (𝑘 + 1) = ((𝑘 + 1) × 𝑘+2 2 ) My math is weak so I dont know what are the proper names to call something