KeypadSDM
B4nn3d
Got a good substitution, reduces the problem to:icycloud said:
cos[t] = Sqrt[(b - a)/(b - x)]
Reduces the problem to integral of:
sec[t]tan[t]dt
Simple...
Reduces the problem to integral of:
sec[t]tan[t]dt
Simple...
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4u Mathematics Marathon V 1.0 (2 Viewers)
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I
icycloud
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Umm nice? I had another substitution in mind though. Let's see the full workingKeypadSDM said:
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KeypadSDM
B4nn3d
Sigh... Here we go:icycloud said:Umm nice? I had another substitution in mind though. Let's see the full working
But now I can see the problem in a greater light. Treat Sqrt[b - x] as the hypotenuse, and Sqrt[x - a] as one of the other sides. Then use sin[t] = Sqrt[(b-a)/(b-x)] And it resolves nicely.
Honestly, can't be stuffed writing the solution I've got, but it's shorter than that other crap one.
I
icycloud
Guest
Yep my method is similar. If you're interested:
I guess we need a new question now.
x = a Cos^2 + b Sin^2
I guess we need a new question now.
I
icycloud
Guest
Post it anyway?Templar said:
You can post two -- one which is ext-2 and the other one which isn't.The question runs on ext 2 combinatorics skills, and is really easy once you see it.
A compositionof an integer is when it is written as the sum of positive integers
n=x1+x2+...+xk, k>0 (ie 3 is a composition of 3)
Assuming that order does matter (ie 1+2 and 2+1 are different compositions of 3), how many compositions are there for integer n (need proper proof/argument instead of a number)?
You can have a crack at this one (and anyone else interested). Someone write a proper ext 2 question.
A compositionof an integer is when it is written as the sum of positive integers
n=x1+x2+...+xk, k>0 (ie 3 is a composition of 3)
Assuming that order does matter (ie 1+2 and 2+1 are different compositions of 3), how many compositions are there for integer n (need proper proof/argument instead of a number)?
You can have a crack at this one (and anyone else interested). Someone write a proper ext 2 question.
I
icycloud
Guest
Templar said:
2^(n-1)? Do I need to prove it? <-- Oops just re-read your question. Working on proof now...
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icycloud
Guest
Templar said:
Hmm I thought it was the n-th row of the Pascal Triangle. Sum (r=0 --> r=n) nCr ???
icycloud said:
Can you show why it is the nth row of the Pascal's triangle? Just making sure you didn't get it by just writing some composition out and looking at a pattern.
And there is a slightly quicker way without using Pascal's.
And there is a slightly quicker way without using Pascal's.
OK someone post an ext 2 question for tomorrow.
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icycloud
Guest
Hehe alright. Still working on it.........Templar said:
An actual ext 2 question based on similar principle from one of the High papers in 2004 (if I remembered the question exactly).
There are 9 rings and 4 fingers. How many ways are there to wear the rings if order on each finger matters and
a. each finger must have at least one ring
b. a finger can have no rings
Keep in factorial/bionomial form.
There are 9 rings and 4 fingers. How many ways are there to wear the rings if order on each finger matters and
a. each finger must have at least one ring
b. a finger can have no rings
Keep in factorial/bionomial form.
SeDaTeD
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Haha, that composition one is really easy. Templar would know what my solution is anyway.
For the fingers and rings.
part a) 9!*8C3
part b) 9!*13C4 Edit: wait, make that 9!*12C3
I'll omit the working out cos i guess Tmeplar would like someone to figure it out.
For the fingers and rings.
part a) 9!*8C3
part b) 9!*13C4 Edit: wait, make that 9!*12C3
I'll omit the working out cos i guess Tmeplar would like someone to figure it out.
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icycloud
Guest
OK, I got a "proof" for the "non ext-2 question"...
For integer n, imagine n boxes, and n-1 separators.
With no separators, obviously there's one way of getting a row of length n. (i.e. 1 + 1 + 1 + ... = n) [(n-1)C0]
With one separator, there are n-1 slots, thus n-1 combinations. [(n-1)C1]
With two separators, there are (n-1)C2 combinations.
...
With n-1 separators, there are (n-1)C(n-1) combinations.
Thus the total number of combinations is:
Sum (r=0 to r=n-1) .. (n-1) C r = 2^(n-1)
OK now what's the proper proof?
With no separators, obviously there's one way of getting a row of length n. (i.e. 1 + 1 + 1 + ... = n) [(n-1)C0]
With one separator, there are n-1 slots, thus n-1 combinations. [(n-1)C1]
With two separators, there are (n-1)C2 combinations.
...
With n-1 separators, there are (n-1)C(n-1) combinations.
Thus the total number of combinations is:
Sum (r=0 to r=n-1) .. (n-1) C r = 2^(n-1)
OK now what's the proper proof?
I was going to leave a note saying you shouldn't spoil the fun with the official solution, but I thought the probability of that happening is low enough not to warrant it.
This reminds me of another question in the same program on L trominos and tiling, solving with induction. It was also taught in first year discrete (which I never turned up) and later appeared in the exam. Later on it was interesting to see all the variations on the methods used to solve it after having done it myself during the exam.
This reminds me of another question in the same program on L trominos and tiling, solving with induction. It was also taught in first year discrete (which I never turned up) and later appeared in the exam. Later on it was interesting to see all the variations on the methods used to solve it after having done it myself during the exam.
KeypadSDM
B4nn3d
Templar said:
For a) just use the divider locations. You've got 9 rings and 3 dividers, with 8 locations for the divider to go in. Since the order of the rings matters, multiply by 9!, hence 9! * 8C3
For b) I was considering doing it by summation, but that seems too long for a simple answer.
Oh wait, now I remember, add in 4 generic rings and add them to each finger, thus the problem degenerates into the first one with 13 rings with only 9 being unique. Hence 13*12*11*...5*12C4 = (13!/4!)*(12C4).
For b) I was considering doing it by summation, but that seems too long for a simple answer.
Oh wait, now I remember, add in 4 generic rings and add them to each finger, thus the problem degenerates into the first one with 13 rings with only 9 being unique. Hence 13*12*11*...5*12C4 = (13!/4!)*(12C4).