jagatuba
  • 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.
Mathematics
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

experimentX
  • 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.
anonymous
  • anonymous
^ Combinations formula. Seems legit.
jagatuba
  • jagatuba
I do understand that, but I want to know how this can be solve with out the use of a calculator.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Factorials?
experimentX
  • experimentX
choosing 'r' out of 'n' can be written as \[ \binom{n}{r} = {n! \over r! (n-r)!} \]
jagatuba
  • jagatuba
Give me another example of what we are talking about that does not involve polygons.
experimentX
  • 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.
jagatuba
  • jagatuba
Very good example: \[\left(\begin{matrix}7 \\5\end{matrix}\right)=\frac{ 7! }{ 5!(7-5)!}\]
experimentX
  • experimentX
yep!!
jagatuba
  • jagatuba
84
jagatuba
  • 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?
experimentX
  • 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?
jagatuba
  • jagatuba
THANK YOU! That is exactly what I was looking for in the original question. Sorry if I wasn't clear.
experimentX
  • experimentX
http://www.wolframalpha.com/input/?i=7+choose+5
jagatuba
  • 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?
experimentX
  • experimentX
lol .. .no without calculator this is horrible.
experimentX
  • 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.
jagatuba
  • 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?"
jagatuba
  • jagatuba
What do you mean the top and bottom cancel out?
experimentX
  • 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.
experimentX
  • experimentX
|dw:1353182495182:dw|
experimentX
  • experimentX
also when 'r' is pretty close to 'n', you can do the same.
jagatuba
  • jagatuba
Oh I see, yes. Still, I'm thinking somewhere I got erroneous information. I'm think that either the CBEST does allow a calculator, OR there are not nearly that many questions that involve factorials. If anyone reading this actually has firsthand experience in taking the CBEST direct message me.

Looking for something else?

Not the answer you are looking for? Search for more explanations.