Problem Solving Using Venn Diagrams

Problem Solving Using Venn Diagrams-81
Data Science Math Skills introduces the core math that data science is built upon, with no extra complexity, introducing unfamiliar ideas and math symbols one-at-a-time.Learners who complete this course will master the vocabulary, notation, concepts, and algebra rules that all data scientists must know before moving on to more advanced material.Since we have three sets – numbers divisible by 3, numbers divisible by 5, and numbers divisible by 7 – we need a three-circle Venn diagram like this: I like to draw these diagrams inside a box so I don’t forget about things that might be floating around outside my sets.

Data Science Math Skills introduces the core math that data science is built upon, with no extra complexity, introducing unfamiliar ideas and math symbols one-at-a-time.Learners who complete this course will master the vocabulary, notation, concepts, and algebra rules that all data scientists must know before moving on to more advanced material.

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Note that we only want the integer part so we’re going to ignore the remainder.

The number of elements in A C = the integer part of (1000/105) = 9 Now we continue to work our way out by looking at the portions of the diagram where two of the three sets overlap: (1) MULTIPLES OF 3 AND 5 (2) MULTIPLES OF 5 AND 7 (3) MULTIPLES OF 3 AND 7 Since we’ve already counted the multiples in the center (the blue part of the diagram) we need to be careful not to count them twice and only find the elements in the red parts of the diagram.

All three circles intersect in the center: This part of the diagram represents the numbers that are divisible by 3, 5 and 7.

If a number is divisible by 3, 5, and 7, it will also be divisible by their least common multiple.

This trick or trap comes up a lot in combinatorics problems.

It’s called “double counting.” To avoid this problem we can use Venn diagrams.Let’s agree the circle labeled A represents all the numbers less than 1000 which are divisible by 3.The circle labeled B represents all the numbers less than 1000 which are divisible by 5.Almost all combinatorics problems can be solved with a single tool that can be used in two different ways – one way when order matters, and another way when order doesn’t matter.However sometime it pays to take a different approach.MULTIPLES OF 3 AND NOT 5 OR 7: The number of elements in A = the integer part of (1000/7) – (9 19 38) = 142 – 66 = 76 Our final diagram: So the number of elements less than 1000 which are divisible by 3, 5 or 7 = 248 96 76 19 57 38 9 = 543 And the number of elements less than 1000 which are NOT divisible by 3, 5, nor 7 = 1000 – 543 = 457 The answer is (A).Keep in mind I’m going to leave you with some challenging problems to try on your own.All of these problems can be solved with Venn diagrams 1. How many positive integers less than 1,000,000 are neither squares nor cubes? Given a random 6-digit integer, what is the probability that the product of the first and last digit is even? No solutions provided, promote peer checking and discussion to justify individual attempts.Taking the right approach is the key to most GMAT and GRE quant problems, and combinatorics problems are no different.

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