[StuR]

I understand the 49C5 would give you the total number of five number combinations but why the 259 again.

I thought the answer would be 49C5/6 because each ticket can hold 6 different five number combinations.

The problem my friend and I encountered was whether it was possible in this number (i.e. whether it was possible to group the combinations into batches of six that fit on one ticket) We thought this is where the problem would occur.

Is there anyway to tackle this? Can you explain the 259 calculation again.

I did something similar with there being 258 (1+6*43)winners (matching at least 5 numbers) for each possible outcome but then came derailed when I said choose one of these 258 for each possible set of 6 numbers = 13.9 million again!!!

Thanks for the input guys.

StuR