If you wish to outline your solution for the benefit of others, that would be great. If not, no worries, i’ll post my methodology tomorrow, unless others do so before hand.
Well, there are 28 ways to pick two items (7 + 6 + 5 …). There are 15 combinations with one of each colour (5 + 5 + 5) I.e. 5 combinations (8-3) for each of the three blue items.
Starting in the top left box, draw a continuous line passing through each box once only, so that the sum of the numbers in each group of four boxes is 24.
So the first four boxes through which the line passes will add up to 24, then the next group of four boxes through which the line passes will also add up to 24 and so on.
You can post your result either by printing the diagram and scanning the resultant diagram with your drawn line. Or simply writing out the numbers in groups of four eg
6,6,3,5:
3,5,8,4: etc (which are clearly not a correct solution).
Variations on a theme are possible using diagonal lines - and they can even cross on the box corners, as that still has only passing once through each box:
I have four possible routes home from work. For the first part of my journey I can either go by train or bus. The probability that I shall go by train is 2/3.
After I get off the train I can either walk or catch a bus. The probability that I shall walk is 3/4.
If the first part of my journey is by bus, then I complete my journey by taxi, with a probability of 1/5, or by walking.
What is the probability that I shall walk part of my way home ?
23/30 (I’m not very confident here, probabilities always confuse me, they don’t immediately show as a pattern to solve in my head, and I only have so many fingers to work them out on.)
Yep Don, same way except I ignored the non-walk paths, so didn’t have the numbers on those, multiplied and then summed in my head three times to be sure I wasn’t missing something.