anonymous
  • anonymous
Evaluate the given binomial coefficient (109/3)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\binom{109}{3}\]?
anonymous
  • anonymous
I have no idea how to do these problems.
anonymous
  • anonymous
Yes, that is it.

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anonymous
  • anonymous
\[\frac{109\times 108\times 107}{3\times 2}\]
anonymous
  • anonymous
else cheat http://www.wolframalpha.com/input/?i=109+choose+3
anonymous
  • anonymous
How does one know which steps to follow to do this problem. Just to confirm, the answer would be the resulting number of that fraction?
anonymous
  • anonymous
yes that is the result
anonymous
  • anonymous
\[\binom{109}{3}=\frac{\overbrace{109\times 108\times 107}^{\text {three terms}}}{3!}\]
anonymous
  • anonymous
So we know to go back three numbers (109, 108, 107) because it's over 3?
anonymous
  • anonymous
Why do we multiply 3 by 2?
anonymous
  • anonymous
the denominator is \(3!=3\times 2\)
anonymous
  • anonymous
as another example \[\binom{10}{4}=\frac{10\times 9\times 8\times 7}{4\times 3\times 2}\]
anonymous
  • anonymous
If the problem were, for instance (50/7), would it be (50x49x48x47x46x45x44/7x6x5x4x3x2)?
anonymous
  • anonymous
yes
anonymous
  • anonymous
it is a whole number in each case, so to compute, cancel first multiply last
anonymous
  • anonymous
And (7/7) would be (7x6x5x4x3x2x1/7x6x5x4x3x2)? Basically I just want to confirm that the denominator just always counts down to 2?
anonymous
  • anonymous
\(\binom{n}{n}=1\)
anonymous
  • anonymous
yes, some people write a 1 there too, but that is silly because multiplying by one is like doing nothing
anonymous
  • anonymous
like they might write \(4!=4\times 3\times 2\times1\)
anonymous
  • anonymous
Gotcha. So the "denominator" will always count down to 2, and the "numerator" will count down the number of the denominator?
anonymous
  • anonymous
doesn't count town to the number in the bottom counts down that many terms (like in the example you wrote)
anonymous
  • anonymous
yeah i guess what you said is right if i interpret it correctly there is also a formula, but i wouldn't use it
anonymous
  • anonymous
Yeah that's what I meant (the number of terms, not all the way down to that number). Thank you very much for explaining this all so clearly!
anonymous
  • anonymous
yw

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