Hm, I don't think that's quite right, but I could be wrong. Here's my formula for the general case (I worked it out as a probability):I think its 42C7 or something
After checking with Wolfram Alpha, I can confirm that my general formula is correct.The way it is to consider writing out the series of integers 1,2,3,4... Imagine placing a red counter over the numbers you select (part of the 9 cards chosen), and a blue counter over the others. Here the question is simplified to 'how many ways are there of arranging 9 blue counters and 41 red counters such that no two blue counters are next to eachother?'. This is a relatively simple question to which the answer is 42C9 = 445891810. I do not know if aa180's general formula is correct, but I think their answer for this specific question is.
Yes, and this formula can be derived quite easily using binary (which is exactly the same method you used to solve this question, but with red and blue counters instead of 1 and 0).There may be something I haven't considered, but that would mean your general formula can be simplified to (N-n+1)C(n)?
Can this be derived in the same way with binary, but grouping them in such a way that you have c 0s behind each 1, then adding up all the individual cases, which are starting with 1, 01, 001 etc?In general, the number of ways of choosing cards from consecutive cards such that there are at least cards between any pair of cards is , which isn't too difficult to prove using double-counting.