STEP I 2012 discussion thread

What's wrong with what he's just posted?

Well he might want to get rid of the text after the solution, but if he's okay with it there then meh

I've no idea what I'm "submitting" to, but sure, take out the "This is what I missed!" bit and the text from "There are 90,000..." to "...3.3616" forms a perfectly valid solution imo. Do with it as you will

Same (but I have no insurance ) Positive thinking and that....AEA'll be fine!



Let me know how you get on on results day - I feel we're very much in the same boat

In 2011 you had to get 47/120 for a 2, so you can't be sure, can you?

Q11(i), I'll add (ii) later.

Goodbye Warwick. The fact this claims to be for just c1-4 is ridiculous as FM helps SO much. The intersection of tangents was an FP3 coordinate system question, differential equation transformations. But still...rant over. Good luck everyone

Done half of question of 9 (by hand), will leave the rest till tomorrow. Good luck STEP IIers

A slight alternative way to think about Q13 is to think about the number of distinct rearrangements for numbers with n number of distinct digits. For example:

There are two forms of 5-digit numbers using 2 distinct digits - aaaab or aaabb.

For digits of the form aaaab, there are 5 choices for a and 4 choices for b. Then the number of distinct rearrangements are (4+1)!/(4!1!) cf. official solution to STEP I 2005 Q1
For digits of the form aaabb, there are 5 choices for a and 4 choices for b. Then the number of distinct rearrangements are (3+2)!/(3!2!)

So total number by using 2 distinct digits is 20*5+20*10 = 300.

You need to be slightly careful in counting the number of choices: for example, all four digit numbers are of the form aabcd. The number of distinct rearrangements is fine - its (2+1+1+1)!/(2!1!1!1!)=60

However there are not 5 choices for a, 4 choices for b, 3 choices for c and thus 2 choices for d. (Why not?) Instead, to count the number of unique combinations, its easiest to proceed as tomp99 did and say there are 5 choices for 'e', the digit that was excluded. Then there are 4 choices for a, which is distinct from b, c and d because there are two 'a's and one of the other digits. So the number of 5-digit numbers using 4 digit numbers is actually 60*5*4 = 1200.


or alternatively :P There are 5C2 ways to arrange the aa. The a gets to be one of 5 digits. Now we want to choose 3 digits from the remaining 4, for the bcd, so we have 4c3, and set b to always b the largest of these digits and c to be the smallest. Clearly the ways to arrange the bcd are 3!, since aa has been placed. this is 5c2*5*4c3*3!=10*4*6*5=1200.


I quite liked tomp99's method of discerning the number of 5-digit numbers with 3 distinct digits - I did it with the above method by considering numbers of the form aaabc and aabbc. Long division at the end is just a bit crap though - they could've been nice and given you a fraction...

However, if you'd done Q1 2005 (and remembered the method), this would've been a nice question to score well on

Right, here is Q5:
$I_2=-2\displaystyle\int^\frac{1}{\sqrt{2}}_1 ulnu\ du$

(The upper limit should be a $\sqrt{2}$ not sure why it is not working)

Copy my LaTex, $\displaystyle I_2=-2 \int^\frac{1}{\sqrt{2}}_1 u \ln u \ du$

The problem probably is due to you not leaving spaces between LaTex codes.