Author Topic: HSC 2010 Probability (Read 760 times) Tweet Share

frog1944 · « **on:** November 01, 2017, 08:46:51 pm »

Hi,

I'm struggling with how to do the HSC 2010 question 7 c) iii) and the reasoning behind it. Could you please explain the solution?

Thanks

Opengangs · « **Reply #1 on:** November 02, 2017, 10:39:25 am »

By expanding the result, we clearly can see that part (b) will be of use.

Part (i) informs us that there are r + 1 combinations that can be made from a selection of r balls.
Part (ii) informs us that there are nCr different selections that can be made.

We can deduce that, for only one particular r, there can be (r+1)(nCr) total number of different selections since for every combination, we have nCr selections.

Thus, to find the total number of different selections, we need to consider cases where r = 0, 1, 2, ... n, which is the sigma notation.
Using part b, we can deduce that:

$\sum_{r=0}^n (r+1)\binom{n}{r} = \sum_{r=0}^n r\binom{n}{r} + \sum_{r=0}^{n} \binom{n}{r}$
$= n2^{n-1} + 2^n = 2^{n-1}(n+2) \text{ (from part b)}$

Search the forums now!

We have moved!

Author Topic: HSC 2010 Probability (Read 760 times) Tweet Share

frog1944

HSC 2010 Probability

Opengangs

Re: HSC 2010 Probability

Recent Posts

User:
Password: