Melody

Answered here:

Where is the pic?

3 sides   angle sum 180 degrees

4 sides   angle sum =  2*180 degrees

n sides   angle sum =  (n-2)*180 degrees

so what is the angle sum of a decagon

Once you know that you just have a simple equation to solve.

Continued here

I have a solution.

I think it is what the other answerer was trying to explain.  (Although I am not sure)

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

Method 1:

The number of subsets of 3 elements from the set of n+2 elements is given by   \(\binom{n+2}{3}\)

Method 2:

Let us  have 2 sets,    A(n elements)  and    B(2 elements)

How many ways are there to choose 3 elements if two MUST come from set B      answer: n

Now take one of the elements from A and put it in set B  so A(n-1 elements)  and B(3 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one    answer: (n-1) 2 = 2(n-1)

Now take one of the elements from A and put it in set B  so A(n-2 elements)  and B(4 elements)

How many ways are there to choose 3 elements if two MUST come from set B but one of the 2 must be the new one    answer: (n-2) 3 = 3(n-2)

....

and so on until all but one of the elements are in set B        n(n-(n-1) = n(1)

So  The number of subsets of 3 elements from the set of n+2 elements is also given by

 \( (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1)\)

HENCE

\(\binom{n + 2}{3} = (1)(n) + (2)(n - 1) + (3)(n - 2) + \dots + (k)(n - k + 1) + \dots + (n)(1),\)

I have shown that they are equal by counting in 2 different ways

If my answer is right then my logic, and therefore my solution,  is also right.

Yes there are ten chairs to choose from but they are all the same. It makes no differnece which is chosen. Take a chair, any chair, and it defines the beginning of the sequence.

I said 1*

you said 10/10 *

They are the same. 

Lets see,

The first stuent has only one place to chose from as they denot the 'begining of the circe'

The second student has 2 places to chose from

The third has 2 to shooe from 

The ninth has 2 to chose from and the last has only one.

1*2*2*2*2*2*2*2*2*1

2^8 = 256 

I am not certain that it is right though.

Please add a link to the previous question/answer

one or two bread from 5    and     any number of fillings from 7   

one type of bread with whatever filling = \(\displaystyle 5\cdot \left[\sum_{n=1}^7\;\binom{7}{n}\right]\)

two types of bread with whatever filling  = \(\displaystyle \binom{5}{2}\cdot \left[\sum_{n=1}^7\;\binom{7}{n}\right]\)

Evaluate and add  

You are very welcome Nerdiest :)