GingerAle

EP probably gave you the point. The “floaters” on the forum annoy him too.

You now have two points.

I saw the solution but I didn't get the insight of the questions and how there were able to derive to formulae... I am new on momentum

Why would you expect such a descriptive reply to a simple, brain-dead question? This question requires only elementary math skills to answer. Posting a nondescript, basic question causes you to appear as a one of the simpleminded idiots, who want someone to do his homework.   

The guest has a point in asking why you are fucking around on here.  Because you are.

Generally, you learn the formulas and their applications before learning the methods of derivation.  The derivation of most physics formulas require advanced levels of calculus, linear algebra, and trigonometry.

If you want an advanced or intermediate level of derivation or the reasons why, then you need to ask pointed questions with contrasting examples.  Here’s an example of such a question.

Most fans of team sports are intuitively aware of the “home court” or “home field” advantage and the disadvantage for away games.  This implies the winning or losing is dependent on the home or away condition.   

Here is a mathematical test for dependent/independent using the chart data.   

\(\text {If this is independent then } \\ P(L|A)=P(L) \tiny \text{ Read as “The probability of (L)osing, given the game is (A)way = the probability of (L)osing. }\\ P(0.3|0.4) \stackrel{?}= P(0.65)\\ 0.75 \neq 0.65 \tiny \text{ These are not equal, so the events are dependent. }\\\)

A notable increase in the probability for losing the away games. 

GA

This is an interesting question.

Here’s a composite graph of the parabolas x=time for y=0 for the back and front of the tuck’s bed, and the corresponding velocities for these range times. 

\(y=5+9t-\dfrac{1}{2}\cdot (9.8)t^2\\ y=7.5+9t-\dfrac{1}{2}\cdot (9.8)t^2\\ y=\text {velocity for time (t) to range target}\\ y=\frac{\left(t\cdot9.81\right)}{2\cdot\sin\left(45\right)}\\\)

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In EP's solution for the front of the truck bed, he adjusted the velocity by constraining the time component. In this case, with the increase in velocity the ball will pass over the front of the truck bed at time = 2.284, but it does not intersect y=0, here.  In range equations, both time and velocity are functions of distance and changing one affects the other.  

The ball with a velocity of 17.4m/s will cross y=0 in (17.4*2*sin(45)/9.81)= 2.51s at (17.4^2/9.8)= 30.86 meters. The target point of the truck will be at 7.5+(9*2.51)= 30.08 meters. So, the ball will land on the hood or the windshield.  

GA

Ginger I expect you have answered some other version of this question.

Yes, I’ve answered many questions related to statistics and combinatorics. I’ve just completed my Masters in Psychology (M.S.) (and Art History (M.A.)). My principle focus is on Abnormal Behavior and Analysis, and this requires a thorough understanding of statistical methods, including combinatorics relating to selection methods. This is why I do well in answering statistical/combinatorics related questions.

My beginning in anything relating to upper-level mathematics was coming to this forum, and particularly meeting Lancelot Link. Before that, I knew only the (rote) probabilities relating to drawing specific poker-hands; I didn’t know how to calculate them mathematically.

GA

More advanced combination/partition set questions will introduce sets with elements that are individually distinguishable. This requires the student-mathamation to discern the relevance.

For this question, the five could all be different or they could be genetic clones; it’s irrelevant to the question. For this question, they are Manoj’s friends and are indistinguishable from each other. 

Of the six friends, only Manoj is distinguishable from the other five. He is distinguishable for the others because he will only attend a class if (and only if) at least one of his friends also attends.  This is why he is separated (distinguishable) from the others. The other individual characteristics of the set of friends are irrelevant.    

By analyzing this question, it’s easy to see the distinguishable fruit is irrelevant here. An orange is the same as an apple. The selection of any fruit creates a fruit basket.

The solution to this is the number of partitions of 15.  

The six (6) friends are assigned to one of two (2) classes. Five (5) are indistinguishable and one (1) is distinguishable. Restrictions: The distinguishable friend must have at least one (1) indistinguishable friend as a classmate. 

Remove the distinguishable friend from the set. Distribute the five (5) indistinguishable to the two classes.  Then assign Manoj only to the classes where one (1) or more of his friends are assigned.

\(\left( \binom{c+f-1}{f}\right) \tiny \textbf{ c=number of distinguishable classes & f=number of indistinguishable friends}\\ \left( \binom{2+5-1}{5}\right) * 2 = 12 \tiny \textbf{ Multiply by 2 because each division creates two sets—one for each class. }\\ \)

These 12 sets will range from zero to five and five to zero. There are two (2) arrangements where a class has zero (0) of Manoj’s friends, leaving 10 class arrangements for Manoj to join.   

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This is also an integer partition of five (5) divided into two sets. Each set represents the two classes with five to zero and zero to five friends.  

{5+0}, {4+1}, {3+2}, {2+3}, {1+4}, {0+5}

The resolution is the same as above.

GA

Mr. BB may have caught a mistake by Sir CPhill, but he didn’t get the solution quite right (no surprises, here).  For one, 14 choose 2 is NOT 90, it’s 91. He didn’t set up an equation for the solution; he solved this intuitively and added some blarney slop, which is his usual method.

The equation is different from the one presented by Sir CPhill. It will look like this:

\(\dfrac{n !}{2!( n-2)!} + \dfrac{n !}{1!(n-1)!} + 1 = 106 \leftarrow \small \text{ solve for n }\\\)

This is much more difficult to solve; it’s an educated guessing process. For small sums, like this one, it’s easy; for large sums, a perusal of Pascal’s triangle would speed up the search, or a computer algorithm.   

Anyway, I eat crow frequently; we chimps like them.  

^{A certain, annoying troll should read his messages more often.}

GA 

This is not the correct solution for this question.

For combinations, a set such as AB = BA, and the two sets are the same and counted only once. If this were the case, then 26C2 would be the correct solution. \(26C2 = \dfrac {26!}{2!(26-2)!} = 325\)

In this question, the order of the sets is unique and counting the permutations of these sets gives the correct solution.

\(26P2 = \frac {26!}{(26n-2)!} = 650\)

Note that the formulas differ only by the division of the set size factorial (2!).

Here’s another method for solving.

Counting all combinations gives \(26^2 = 676\) sets of two letters including AA , BB, ect.

The double letters occurs once for each of the 26 letters. Subtracting the 26 duplicates from 676 leaves 650 sets, and this the number of sets without a repeated letter.

Note also that alphabetizing does not make a difference in the set counts. 

GA