Please Help...Maths is hard.

[Flavus Aquila]

Damn, that's what I thought it was at first but then thought it was a double trick question and second guessed myself.

(n/2)x(n-1) is correct but i'm not sure why you divide by two. This is pretty much the useless part of math anyways. It's all about calculus, even though it can make you its b***h.

[MENTION=3545]bickelz[/MENTION];
Why divide it by two?

If you have a set of eight people shaking hands, you can add the handshakes up: 7+6+5+4+3+2+1 =28 handshakes.
But you can notice that if you add up the first and last number of such a series, you get the same number if you add up the second and second last of such a series: ie: 7+1 = 8 ; 6+2 = 8; 5+3 = 8; etc. The four just sits on its own. Two things are apparent then, if you average out each number in a series, for example 7+1 = 8/2 = 4; that firstly the average of each number in a series is half the original number; and secondly, you don't need to actually add up a descending series of numbers to get the number of handshakes. You just need to multiply the average total of each number in a series by the length of the series.

So n/2 gives you the average value of each number in a descending series ie. 8/2 = 4.
And since in a group of 8, one person cannot shake his own hand, n-1 gives you the length of the series ie. 8-1 = 7.
Now you just have to multiply the average value of each step, by the number of steps: (n/2)(n-1): 4*7 = 28.

 

[Orion]

Also, different teachers approach mathematics differently. Some kids learn very effectively with the rote method but others need to understand the "why", the underlying dynamics, the patterns as [MENTION=1678]Norton[/MENTION] said.

This is pretty much why I zoned out of all maths, science, geography.. *insert cold, uninteresting subject* classes.

The teacher could never give me the "why". Why is this important? How does this fit into the context, the grand scheme of things? The thing is, I look back now and think that all those classes are now extremely useful to me because I have a different opinion of learning now and how everything interrelates. But I'm only interested now in maths, physiology, biology etc because I have a context for it and I understand why it's important to my main interests.

 

[Arsal]

number of handshakes = (n-1)+(n-2)+(n-3)+...+(n-n)

 

[aerosol]

number of handshakes = (n-1)+(n-2)+(n-3)+...+(n-n)

I'm going to rep this based on how fancy it looks.
I have no idea if it's correct because I'm not into math. Me and math have a strained relationship. I understand it and its good qualities but it is so boring to be around and we never have any fun together.



But it's Arsal, so I'd put money on it.

 

[Arsal]

I very much suck at Maths, so I wouldn't put my money on it. For one thing, (n-n) is incorrect, because it cancels itself so writing it again is redundant. But I didn't know how else to denote the last element in the series, given that "summation of (n-i)" in notation wasn't possible without graphical support, which would be the accurate representation of what I wanted to say.

 

[aerosol]

I very much suck at Maths, so I wouldn't put my money on it. For one thing, (n-n) is incorrect, because it cancels itself so writing it again is redundant. But I didn't know how else to denote the last element in the series, given that "summation of (n-i)" in notation wasn't possible without graphical support, which would be the accurate representation of what I wanted to say.

I edited and deleted thisssss space.

Arsal was too embarrassed.

 

[Arsal]

maybe not...

 

[daydreamer]

Math is mah bitch.

Then tell your bitch to please be nicer >.>
I'm not doing well in math right now
haha

 

[bickelz]

@bickelz;
Why divide it by two?

If you have a set of eight people shaking hands, you can add the handshakes up: 7+6+5+4+3+2+1 =28 handshakes.
But you can notice that if you add up the first and last number of such a series, you get the same number if you add up the second and second last of such a series: ie: 7+1 = 8 ; 6+2 = 8; 5+3 = 8; etc. The four just sits on its own. Two things are apparent then, if you average out each number in a series, for example 7+1 = 8/2 = 4; that firstly the average of each number in a series is half the original number; and secondly, you don't need to actually add up a descending series of numbers to get the number of handshakes. You just need to multiply the average total of each number in a series by the length of the series.

So n/2 gives you the average value of each number in a descending series ie. 8/2 = 4.
And since in a group of 8, one person cannot shake his own hand, n-1 gives you the length of the series ie. 8-1 = 7.
Now you just have to multiply the average value of each step, by the number of steps: (n/2)(n-1): 4*7 = 28.

That makes sense but I don't like using stuff like this unless I can see the derivation of the equation. It's more logical than theoretical.

 

[frozen_water]

There's a more formal way to do it, although it's a little bit trickier to understand.

The formula you're trying to show is that 1 + 2 + ... + n = n*(n+1)/2

So first you do something simple. Let n = 2. In that case, 1 + 2 = 3. The other half of the equation goes:

n*(n+1)/2
2*(2+1)/2
6/2
3

so this formula works in the case where n=2.

Then you want to prove: "if it works for the number 'n', then it works for the number 'n + 1'"

in math language, the "if" part of that statement is the same as saying that 1 + 2 + 3 + ... + n = n*(n+1)/2

So we assume 1 + 2 + 3 + ... + n = n*(n+1)/2, and then just do algebra:

first add (n + 1) to both sides, to get

1 + 2 + 3 + ... + n + (n+1) = n*(n+1)/2 + (n+1)

Then by manipulating the right side, we get:

n*(n+1)/2 + (n+1) (same as above)
(n^2 + n)/2 + (n + 1) (multiply n*(n+1))
(1/2)*n^2 + (1/2)*n + n + 1 (distribute the division-by-2)
(1/2)*n^2 + (3/2)*n + 1 (add (1/2)*n + n)
(1/2) * (n^2 + 3*n + 2) (pull a factor of (1/2) out of each term)
(1/2) * (n+2) * (n+1) (factor n^2 + 3*n + 2)
(n+1)*(n+2)/2 (rewriting "1/2*" as "/2")

so if we assume that 1 + 2 + ... + n = (n)*(n+1)/2,
we can prove that 1 + 2 + ... + n + (n+1) = (n+1)*(n+2)/2

but since we showed that it works when n = 2, that means that our assumption is true in at least this case.

But if it's true when n = 2, then we can prove that it's true when n = 3.
and if it's true when n = 3, then we can prove that it's true when n = 4.
and if it's true when n = 4, then we can prove that it's true when n = 5.

...etc.

That's the more "rigorous" way to prove that the formula works all the time, if that makes you more comfortable.

 

[Artsu Tharaz]

Each of the 25 people shakes the hands of 24 other people. That's 25*24 handshakes.

However, this means we are counting handshakes from the perspective of each person - essentially, counting the number times a single hand shaked. But a handshake requires two hands, i.e. we aren't supposed to count it once from person A's perspective and another time from person B's perspective.

We therefore counted each handshake twice, so we have to halve our result, i.e. (1/2)*25*24