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Mathematics 12 Online
OpenStudy (jagatuba):

I know that 35 unique quadrilaterals can be formed from a heptagon by joining the vertices. I can work that out in a diagram. What I want to know is the math behind figuring this out. An equation that will work with any shape (octagon for instance), and/or how this equation is derived. Detailed explanation please.

OpenStudy (experimentx):

http://www.wolframalpha.com/input/?i=7+choose+4 4 vertices make quadrilaterals. Heptagon has 7 vertices. So what you are doing is choosing 4 vertices out of 7 in unique ways.

OpenStudy (anonymous):

^ Combinations formula. Seems legit.

OpenStudy (jagatuba):

I do understand that, but I want to know how this can be solve with out the use of a calculator.

OpenStudy (anonymous):

Factorials?

OpenStudy (experimentx):

choosing 'r' out of 'n' can be written as \[ \binom{n}{r} = {n! \over r! (n-r)!} \]

OpenStudy (jagatuba):

Give me another example of what we are talking about that does not involve polygons.

OpenStudy (experimentx):

most common example:- In a classroom you have 7 students. You have choose 5 students for basketball team. In how many different ways can you choose students.

OpenStudy (jagatuba):

Very good example: \[\left(\begin{matrix}7 \\5\end{matrix}\right)=\frac{ 7! }{ 5!(7-5)!}\]

OpenStudy (experimentx):

yep!!

OpenStudy (jagatuba):

84

OpenStudy (jagatuba):

Now is there a quicker way to figure factorials without the use of a calculator? Obviously, 15! is going to take time and be cumbersome to figure out manually. Is there a short cut?

OpenStudy (experimentx):

the topic is "Permutation and Combination" Permutation = choose something in order Combination = choose but order does not matter. ----------------------------------------- like you have two benches and 10 students. Each bench can hold 5 students. Q1. In how many ways can students sit in first bench? Q2. In how many ways can you ARRANGE student in first bench?

OpenStudy (jagatuba):

THANK YOU! That is exactly what I was looking for in the original question. Sorry if I wasn't clear.

OpenStudy (experimentx):

http://www.wolframalpha.com/input/?i=7+choose+5

OpenStudy (jagatuba):

But seriously. If you have to do some extensive factorial work without a calculator, is there a short cut to going for example; 15*14*13*12* . . . *3*2?

OpenStudy (experimentx):

lol .. .no without calculator this is horrible.

OpenStudy (experimentx):

usually in combinations, the top and bottom cancel out so that it makes thing a bit simpler. that's the only easy portion when dealing with large numbers.

OpenStudy (jagatuba):

That's what I thought. See the thing is I'm studying for my CBEST and I was told that you cannot use a calculator, but I was just taking a sample test online and there are all these factorial questions that I'm like "look, I don't want to cheat, but how am I supposed to manually figure out all of these without a calculator in the allotted amount of time?"

OpenStudy (jagatuba):

What do you mean the top and bottom cancel out?

OpenStudy (experimentx):

sorry .. numerator and denominator. when 'r' is close to 'n'. you can cancel out numerators and denominators and figure it out easily without calc.

OpenStudy (experimentx):

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