Mega A Level Maths Thread MK II

No logs

I used that! YAY! and here's what I've got so far:

$\frac{n-1!}{r-1!(n-1-(r-1))!} + \frac{n+1!}{0!(n+1)-0)!}$ I know it can be simplified but I'm wary about the 0...I think its meant to be 1...

There are no points of inflection:
as $\dfrac{d^2y}{dx^2} \not=0$
and nothing to intersect with

0! = 1

I used that! YAY! and here's what I've got so far:

$\frac{n-1!}{r-1!(n-1-(r-1))!} + \frac{n+1!}{0!(n+1)-0)!}$ I know it can be simplified but I'm wary about the 0...I think its meant to be 1...

I see.
Seeing as there's some talk of IMO, I thought I'd ask "have you ever participated in a mathematics competition?". If so, how did it go?

Qualified for BMO1 this year. Didn't get to BMO2 but I'm hell bent to next year

0! = 1

BMO2>STEP right?

Lets go for something a bit different.
Show that:
$^{n-1}\mathrm{C}_{r-1} + ^{n-1}\mathrm{C}_{r}=^n\mathrm{C}_r$

It should read:

I've always found that a little hard to grasp.

I would've thought the level of problem solving they learn from the IMO is transferable...I've heard BMO2's harder than STEP ...

blub blurb ... right I got part 1 right, the numerator on the top half of the second part was a silly mistake, how did you get the denominator?



Yeah.. its a bit dodge, how does that even work out

Because, I think: $n! =\dfrac{(n+1)!}{(n+1)}$

I've always found that a little hard to grasp.

Yeah.. its a bit dodge, how does that even work out

Hi



For the first one, the word THURSDAY has 8 letters, and two vowels (U and A).

Four letters are picked at random.

The number of possible combinations of 4 letters you can get from THURSDAY would be 8C4 = 70.

Now, they want the probability that there is at least one vowel among the letters.

There are 6 consonants/non-vowels in THURSDAY - T, H, R, S, D, Y.

So the number of possible combinations of 4 letters you can get from the consonants alone would be 6C4 = 15.

So the probability that you'd have NO vowels would be $\dfrac{15}{70} = \dfrac{3}{14}$

And thus the probability that you'd have at least one vowel would be $1 - \dfrac{3}{14} = \dfrac{11}{14}$



For the second one, for the number to be greater than 4000, the first digit would have to be 5, 6 or 8.

If the number starts with 5, 6 or 8, we'd have 4 numbers left.

The number of permutations of these 4 numbers in the 3 spots left would be 4P3 = 24.

So we'd have 24 numbers starting with 8, 24 starting with 6, 24 starting with 5.

So the total would be 24+24+24 = 72

ahh $\dfrac{d}{w}$ I've gotten it now, thank you anyways!

It comes from the gamma function:

$\Gamma (z) = (z-1)! = \displaystyle\int_{0}^{\infty} t^{z-1} e^{-t} dt \Rightarrow 0! = \Gamma(1) = \displaystyle\int_{0}^{\infty} e^{-t} dt = \left[ - e^{-t} \right]_{0}^{\infty} = 1$

At least I think. Where are you L'art?

This is not completely right, because there are 5 numbers then any 5 digit no. will be in the 10,000's ie above 4000;
Therefore, there are 4! (or 4p4) ways of arranging the numbers outside of the ten thousands (n) ie. n - - - -
and because any value can be used for n there are 4! x 5 no.s bigger than 4000;

total = 72 (from previous question) + 120 = 192

hope that made sense!

Can someone help me with C1 Solomon paper c (5^4/9)^-1/2

How do I do this?