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I don't see where you could have went wrong but 2640 is incorrect

I suspect the following is what is required:

Hmm. I understand. I tried 2640 but that is incorrect? 

Yes, at the end of the day, that is what we are counting for, but we count all the ways to "fail" and subtract that from the total ways. 

For example, here's a simpler problem: You flip a fair coin twice. How many ways can you get at least 1 head?

For this problem, instead of counting the ways we can get a head, we count all the ways we don't get a head. This can happen in 1 way (back-to-back tails), and there are 2 x 2 = 4 total ways to flip the coin, so there are \(4 - 1 = 3\) ways to get at least 1 head. 

We are doing the same thing here, just that it is a bit more complicated. If we subtract all the failures(where Richard does not sit next to S or T) from the total ways to seat the 8 people, we are left with the number of ways where Richard sits next to S or T. 

But we want richard to sit next to surya and or tamas

We are using complementary counting, which means we count for everything we don't want. 

In this case, we are counting all the ways we can have Richard not sit next to S or T, so we restrict the number of people who can sit next to Richard. 

Hope this helps!

Here we go again... my computer froze, so now I need to retype everything

Let a and b represent the number of students in divisions A and B, respectively. 

There are \(3 \times {a \choose 2} = {3a(a-1) \over 2}\) ways to pair the teams in Division A, and \(2 \times {b \choose 2} = b(b-1)\) ways to pair the teams in Division B.

There are also \(ab \) ways to pair the students from different divisions. 

This means we can form the equation \({3a(a-1) \over 2} + b(b-1) + ab = 454 \ \ \ \ (i)\)  

We also know that \(a + b = 23 \ \ \ \ (ii)\)

Solving the system and choosing the integer solutions gives us, in terms of \((a,b)\), \(\color{brown}\boxed{(13, 10)}\)

Help pls fast

Why cannot it not be S or T?

There are \(8! \div 8 = 5040 \) ways to seat the 8 people (divide by 8 to account for rotations). 

Let's use complementary counting for this problem, as it avoids counting multiple cases. So, we are looking to find all the failures (situations where Richard isn't sitting next to either person)

There are 8 spots for Richard to sit in, 5 spots for the person to the left of Richard (can't be S, T, or R), and 4 spots for the person to the right of Richard (can't be S, T, R, or the person to the left of Richard). Lastly, there are \(5! = 120\) ways to seat the other 5. However, we have to divide by 8 (to account for rotations), so there are \(8 \times 5 \times 4 \times 5! \div 8 = 2400\) ways to "fail". 

This means that there are \(5040 - 2400 = \color{brown}\boxed{2640}\) successes. 

Thank you both!

Hi Alan, nice seeing you here!

The question is written in an ambiguous fashion, and it is unclear what was meant. 

There are \({20 \choose 2} = 190\) ways to choose 2 people to shake hands. 

Of these, there are 10 ways to shake hands with the same team member, so there are \(190 - 10 = \color{brown}\boxed{180}\)

There are 2 cases to consider: \(a = b + 1 \) and \(b = a + 1\)

So, we have 2 systems:

\(3a - 4b = 24\)

\(a = b + 1\)

AND
 

\(3a - 4b = 24\)

\(b = a + 1\)

The solutions for the two systems, respectively in the form of \((a,b)\), are \(\color{brown}\boxed{-20, -21}\) and \(\color{brown}\boxed{-28, -27}\)

The first   team  will have a total of   18 + 18 = 2*18  unique  handshakes

The second team will have a total of  16 + 16 =  2*16  unique handshakes

The third team  will  have a total  of 14 + 14  = 2*14 unique handshakes

.

.

The ninth team  has a total of   2 + 2 = 2 * 2  unique handshakes

So the total =  2 ( sum of the first 9 even positive integers)  =

2  (n) (n + 1)    =

2 (9) (10)  = 

180 handshakes

Suppose 3b-4a = 24. Given that a and b are consecutive integers, and b

Maybe try typing your question without LaTex

I'm trying to make it work but it is glitching.

"and b

Are you sure your question is formatted properly?

It just ends with ",and b"...

Thanks so much in advance!!!

LOL, the formatting is off. The unfinished part is "and b

Let the triangle be ABC

We can find the area of ABC thusly

Semi-perimeter  =   (9 + 8 + 7)  / 2    =  12

Area  =   sqrt [  12 (12 -9)(12 -8) (12-7) ] =  sqrt [ 12 * 3 * 4 * 5 ] =  12sqrt (5)

The height can  be figured as

12sqrt (5)   =  (1/2) (8) (h)

(3)sqrt (5)  = h

Line CD  divides the perimeter in half

Line CE divides the area in  half

CF  =  the height

Angle EDC is the acute angle (theta)  between  these two segments

 angle  EDC  =  angle DCF  - angle ECF

tan   DCF  = DF / CF  =  3 / ( 3sqrt (5))    =  1/sqrt (5)

tan  ECF  = Ef/ CF =   2 / (3sqrt (5) )

tan (EDC)  = tan [ DCF - ECF]  =   [  tan DCF - tan ECF ]  /  [ 1 + tan(DCF)* tan(ECF] ] =

tan (EDC)  =   [ 1/sqrt (5)  - 2/ (3sqrt 5 ) ]  /  [ 1 + (1/sqrt 5)(2/(3sqrt 5)) ] =

[ 1 / (3sqrt 5 ]  / [  1 + 2 / 15]  =

15  / [  17 * 3 sqrt 5 ]  =

15 / (51*sqrt 5 )  =

(5 / (17*sqrt 5) =

sqrt (5)  / 17

10101101

A==Apples,    G==Oranges

A  -  3 == 1 / 4 G,

G  -  3 ==A, solve for A, G

A ==5 apples

G ==8 oranges

5  +  8 ==13 - total fruits altogether. 

I think N($3,456, $478) is meant to indicate a normal distribution with mean $3,456 and standard deviation $478 (or possibly a variance of $478, as both forms are common, and the poster hasn't specified which).