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jagatuba

  • 2 years ago

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.

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  1. experimentX
    • 2 years ago
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    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.

  2. CliffSedge
    • 2 years ago
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    ^ Combinations formula. Seems legit.

  3. jagatuba
    • 2 years ago
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    I do understand that, but I want to know how this can be solve with out the use of a calculator.

  4. CliffSedge
    • 2 years ago
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    Factorials?

  5. experimentX
    • 2 years ago
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    choosing 'r' out of 'n' can be written as \[ \binom{n}{r} = {n! \over r! (n-r)!} \]

  6. jagatuba
    • 2 years ago
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    Give me another example of what we are talking about that does not involve polygons.

  7. experimentX
    • 2 years ago
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    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.

  8. jagatuba
    • 2 years ago
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    Very good example: \[\left(\begin{matrix}7 \\5\end{matrix}\right)=\frac{ 7! }{ 5!(7-5)!}\]

  9. experimentX
    • 2 years ago
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    yep!!

  10. jagatuba
    • 2 years ago
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    84

  11. jagatuba
    • 2 years ago
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    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?

  12. experimentX
    • 2 years ago
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    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?

  13. jagatuba
    • 2 years ago
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    THANK YOU! That is exactly what I was looking for in the original question. Sorry if I wasn't clear.

  14. experimentX
    • 2 years ago
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    http://www.wolframalpha.com/input/?i=7+choose+5

  15. jagatuba
    • 2 years ago
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    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?

  16. experimentX
    • 2 years ago
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    lol .. .no without calculator this is horrible.

  17. experimentX
    • 2 years ago
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    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.

  18. jagatuba
    • 2 years ago
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    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?"

  19. jagatuba
    • 2 years ago
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    What do you mean the top and bottom cancel out?

  20. experimentX
    • 2 years ago
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    sorry .. numerator and denominator. when 'r' is close to 'n'. you can cancel out numerators and denominators and figure it out easily without calc.

  21. experimentX
    • 2 years ago
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    |dw:1353182495182:dw|

  22. experimentX
    • 2 years ago
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    also when 'r' is pretty close to 'n', you can do the same.

  23. jagatuba
    • 2 years ago
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    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.

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