MAT Prep Thread - 2nd November 2016

In the case where you choose the same for both, you don't end up double counting, so dividing the complete total ends up giving the wrong answer.

E.g. with 3 sweet types A, B, C the options including order are:

AA, AB, AC
BA, BB, BC
CA, CB, CC

AA, BB and CC only appear once, it's only the cases where we have 2 different letters that we end up double counting (i.e. AB+BA, AC+CA, BC+CB).

Thats what I thought too which is why I got 210 instead of 200 for 20 possibilities?

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Would g(a,b) = s( b-1 , p(a) , p(f(a,m(b)) work for 2015 Q5 part IV?

The idea being that its one less than f(a,b) which = a + b + 1 as g(a,b) has one less iteration due to x = b-1 instead of x=b

Did you miss that part (iv) asks for a formula valid for $b \leq 0$?

Did you miss that part (iv) asks for a formula valid for $b \leq 0$?

Thats what I thought too which is why I got 210 instead of 200 for 20 possibilities?

Posted from TSR Mobile

Your formula also doesn't satisfy the requirement that it uses only the functions p, m and s (because it uses the function f).

The details of whether you can "just" use your formula but replace b with -ve b depends a bit on exactly how you resolve this. But I'd say "just" isn't going to be a fair description of the process...

EDIT: TBH, I think the easiest thing is going to be to look at your explanation in (iii), and therefore see how you can craft your own function g using the same ideas.

2nd EDIT: Just in terms of the "intent" of the question; the general idea is about how to use some really simple operations (adding 1, subtracting 1, choosing between items based on if something is >= 0) to build more complex operations that let you add numbers together. In which case using "-b" in your answer is "cheating", because you've brought in an operation (function) beyond the ones described. (Explicitly, the negation of b is just as much a function as p, m and s). I think it would be harsh to be penalised for this at this level, but I'm sure it's not the desired approach.

I ended up going for g(a,b) = s( p(b) , p [g ( a , p(b) ) ], p(a) )

Not sure if this would work, the mark scheme has a similar, but slightly different answer.

for 2009 Q2 iv why do they only add up to n-1 instead of n terms?

Hmm?

Am I being stupid again?

No no, i'm just not sure what you're asking, it's me being stupid haha

Well from what I can see they have left the 1 added the 2, 4, ...2n
by the formula n/2(first term + last term) but instead of having n it is upto n-1?

If you're referring to the number of terms, because they're not including the 1 there are n-1 terms when going up to Yn.

Yeah but they aren't taking the one into account anyway no? Because they used 4 as the first term