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You can get those substitutions by letting a+b=x, b+c=y and a+c=z, then solving simultaneously for a, b and c.

I derived that as an example of something you could derive. It was not designed to be quoted.

Of course, we are really dividing into groups of 3, 4, 5, 6 and 7.

Re: 2015 permutation X2 marathon Yes, but you need to explain why they follow the Fibonacci sequence. Seeing that the pattern exists is only the first step.

Re: 2015 permutation X2 marathon A subset is any group of any size selected from the original set. That includes the empty set (size zero) and the entire original set. You should already be able to calculate the total number of subsets that can be formed from a set of 12 objects. But...

Re: 2015 permutation X2 marathon You can have any set of just one number. You can't have {2,3}, {2,3,4} or {2,4,6,8,10,11}. You can have {2,4}, {2,4,6,8,10,12}, {1, 9, 12}.

Re: 2015 permutation X2 marathon Sorry - forgot. Edited now.

Re: 2015 permutation X2 marathon Including the empty set, how many subsets of {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12} do not contain any consecutive integers? Answer: 377 (Since I am not going to count what question we are up to, perhaps we should mark new questions as I have - in the...

Re: 2015 permutation X2 marathon Are you sure about your answer? I am getting 10 more.

Re: 2015 permutation X2 marathon Because your initial move is 1 step OR 2 steps. There are S(n-1) moves you can make if your first step is 1, and another S(n-2) moves you can make if your first step is 2.

Re: 2015 permutation X2 marathon Let S(n) be the number of ways of climbing n steps. The first move is either 1 step or 2 steps. If it is 1 step, there are S(n-1) ways of climbing the remaining n-1 steps. If it is 2 steps, there are S(n-2) ways of climbing the remaining n-2 steps. So...

Re: 2015 permutation X2 marathon 13th

Re: 2015 permutation X2 marathon So no-one can see the connection to Fibonacci numbers?

In Ext 2 last year, only 1st place scored 100 (at least I'm assuming he got 100).

Re: 2015 permutation X2 marathon Without referring to nCr, nPr, factorials, or even simple multiplication, how else can this answer be generated? (Think Robert Langdon) [BTW, not sure what 'stars and bars' refers to, but you can get your calculation by simply counting "words" that can be...

Re: 2015 permutation X2 marathon In how many ways can you climb 12 stairs by stepping 1 or 2 steps at a time? (Not necessarily all 1's or all 2's) Answer: 233

Re: 2015 permutation X2 marathon It is a LOT easier than that. Each man sits next to 2 of the 5 women. So the probability is 2/5.

Re: 2015 permutation X2 marathon Of course, there would be nothing wrong if that were not the case, right?

Re: 2015 permutation X2 marathon I had 6/10C2 and somehow thought 10C2 was 90. Still don't know though why people insist on applying ordering to questions that don't require ordering.

Re: 2015 permutation X2 marathon If the 2 people randomly choose seats, the probability of them choosing two adjacent seats on the same side of the table is definitely 1/15 regardless of whether sides are considered identical or different. I'm not sure what logic you are using to get that answer.