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\(\dbinom{16}{5}=\dfrac{16\times 15\times 14\times 13\times 12}{5\times 4\times 3\times 2\times 1}=\boxed{4368}\)

And that's how it's done.

Very nice, Mr. O   !!!!!!

Here is the full answer:

\(\begin{align*} \dbinom{15}{2} &= \dfrac{15!}{13!2!} \\ &= \dfrac{15\times 14}{2\times 1} \\ &= 15 \times \dfrac{14}{2} \\ &= 15 \times 7 \\ &= \boxed{105}. \end{align*}\)

15nCr2 =105

85nCr82 =98,770

We have 

[a₁ + a_n ] ( [ a_n - a₁ ] / 3 + 1 ) / 2

Where a₁ and a_n are the first and last terms to be summed  and 

( [ a_n - a₁ ] / 3 + 1 ) / 2     are the number of pairs of equal sums to be added together

So we have 

[ 14 + 317 ] ( [317 - 14 ] / 3  + 1 )  / 2  =  16881

Number of terms = [317 - 14] / 3  +  1 = 102

[F + L] / 2  x 102 =, where F=First, L=Last.

[14 + 317] / 2  x 102 =16,881

This is equal to \(14+(14+3)+(14+3*2)...\)

Separate the 14s: \(14*102+3+3*2+...+3*101\)

Factor: \(14*102+3(1+2+...+101)\)

Evaluate: \(1428+3(1+2...+101)\)

Sum of consecutive integers = \({n(n+1)\over2}\)

Evaluate again: \(1428+3({101(102)\over2})\)

\(1428+3*5151\)

\(1428+15453\)

Final answer = \(16881\) 

My interpretation: The shortest side of the right triangle is 6cm and the hypotenuse is 18cm.

Pythagorean Theorem: \(leg1^2+leg2^2=hyp^2\)

\(6^2+?^2=18^2\)

\(36+?^2=324\)

\(?^2=288\)

\(?=\sqrt{288}\)

Simplifying, we get \(?=12\sqrt{2}\) cm. 

Sorry, yes there was, but I got the problem now. Thanks!

\({x \choose y}=\frac{x!}{y!(x-y)!}\)

Knowing this formula will allow you to compute any input for the choose function. Now, let's compute the result.

A Perfect number is a number that is = to the sum of its proper divisors.

Example: 6 = 1 + 2 + 3.

Find the areas of the triangles around it (which is 4 units) and subtract it from the entire square (9 units)

OR

Find the areas of each traingle inside the square and add 1 unit for the square by itself in the middle

16C15 = 16   when you get more familiar with this the answer will become obvious to you.

It is like saying how many ways can 15 obects be selected from 16 (order doesn't count)

and this is the same as saying how many ways can one object be omitted from the original 16. There is 16 objects so there are obviously 16 ways that one can be left out.

16Cr = 16C(16-r)   it is always symmetrical like this. 

There are different ways to enter this on the web2 calc 

I would do it like this (if I had needed to)

nCr(16,15)=

nCr(16,15) = 16

7nCr6 =7

Ahhh, I got the answer:

\(\dbinom{16}{15}=\dbinom{16}{1}=\boxed{16}.\)

"The mass of a dust particle is approximately 7.5x10^-10 kilograms

and the mass of an electron is 9.1x10^-31 kilograms.

Approximately how many electrons have the same mass as one dust particle?"

This is just a division problem. You want electrons are in each dust particle

So how many times 9.1x10^-31 kilograms. goes into 7.5x10^-10 kilograms

7.5x10^-10 kilograms      divided by     9.1x10^-31 kilograms 

(7.5x10^-10) / (9.1x10^-31)  

=(7.5/9.1) * (10^-10 / 10^-31) electrons per particle

(7.5/9.1) = 0.8241758241758242 = 8.241758241758242*10^-1

(10^-10 / 10^-31) = 10^21

So tha answer is   

\(\approx 8.24\times 10^{-1}\times 10^{21}\\ \approx 8.24\times 10^{20}\;\;\text {electons in one dust particle}\)

Just divide one by the other. Why did you ask your question in the "Answers" section instead of in the "Questions" section?

(7.5 x 10^-10) / (9.1 x 10-31) =8.24 x 10^20 electrons in 1 mote of dust.

Mr. Owl, is there supposed to be something more to the sentence

". . . the ratio of participants in debate, music, and community service, respectively, is."

???  

"The mass of a dust particle is approximately 7.5x10^-10 kilograms and the mass of an electron is 9.1x10^-31 kilograms. Approximately how many electrons have the same mass as one dust particle?"

(X stands for multiplication in this case)

Ok, shoot, and I will see what I can do for you. (I am certified to tutor math for Kindergarten- 4th grade.) But I do enjoy higher grade math challenges! 

Can you define n for me? Then I can solve it.