Almintas2l
2022-07-17
Answered

Discrete Math Combination Question. How many ways are there to pick a five-person basketball team from 12 possible players? How many selections include the weakest and the strongest players?

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bardalhg

Answered 2022-07-18
Author has **15** answers

Step 1

If the 12 players are subdivided into 6 strongest and 6 weakest, then the number of ways to select a five person team including players from both subsets is counted by excluding the complement: we take the total count then subtract the count of ways to pick all players just from either category.

Step 2

$(}\genfrac{}{}{0ex}{}{12}{5}{\textstyle )}-2\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{6}{5}{\textstyle )$

If the 12 players are subdivided into 6 strongest and 6 weakest, then the number of ways to select a five person team including players from both subsets is counted by excluding the complement: we take the total count then subtract the count of ways to pick all players just from either category.

Step 2

$(}\genfrac{}{}{0ex}{}{12}{5}{\textstyle )}-2\phantom{\rule{thickmathspace}{0ex}}{\textstyle (}\genfrac{}{}{0ex}{}{6}{5}{\textstyle )$

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I am working on a homework assignment and I am having trouble understanding the problem. I feel as if my professor forgot part of the problem, but I would just like to double check and make sure I am not reading the problem incorrectly. This is the problem:

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So essentially ${5}^{x}\equiv 8\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}13)$

However, In this problem 5 is not a generator of ${\mathbb{Z}}_{13\cdot}$ 5 produces a cyclical group {5,12,8,1} which does not contain all the elements of ${\mathbb{Z}}_{13\cdot}$

I understand there is a solution to this, but can someone clarify this rule for me?

${\mathbb{Z}}_{p}$ under multiplication, $p=13,Z=\{1,\dots 12\}$ I need to find the discrete logarithm of base 5 to 8:

So essentially ${5}^{x}\equiv 8\phantom{\rule{0.444em}{0ex}}(\mathrm{mod}\phantom{\rule{0.333em}{0ex}}13)$

However, In this problem 5 is not a generator of ${\mathbb{Z}}_{13\cdot}$ 5 produces a cyclical group {5,12,8,1} which does not contain all the elements of ${\mathbb{Z}}_{13\cdot}$

I understand there is a solution to this, but can someone clarify this rule for me?

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This is an example question from my Discrete Mathematics course textbook. I'd like help on whether I'm approaching it correctly and how to continue from where I am, that is if I'm right so far.

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This is an example question from my Discrete Mathematics course textbook. I'd like help on whether I'm approaching it correctly and how to continue from where I am, that is if I'm right so far.

A coin is flipped six times where each flip comes up either heads or tails. In how many possible outcomes are the number of heads and tails not equal?

I've figured out that order doesn't matter, indicating I'll use combination and that there are 64 possible outcomes $({2}^{6}=64)$ and that for the number of heads and tails not to be equal any outcome is possible apart from 3 heads and 3 tails. I'm fairly new to this so I'm not sure how to proceed from here.