I got 42C9 using the method I posted above, so I assumed there was a simpler method.Thanks, but i think this is way over the top for 4unit.

I saw a way a teacher did it through binary. The analogy goes something like this:

1 0 1 1 1 1 0 0 0 is a binary string of length 10. In a binary string of length 50, how many ways are there to have a string with exactly nine lots of ones and that no ones are next to each other?

I fail to see how these two are the same question/where is the one-to-one correspondence.

Maybe trebla can help?

For the equivalence between the method your teacher used and the original question, it is as follows:

The bit string of length 50 corresponds to your choice of whether you pick each number from 1-50 or not. A bit of 0 in the string means you did NOT pick that number, and a bit of 1 means you picked that number.

For example, (assume you're picking 3 numbers from a 1-10 for simplicity here): 1001000100 has a 1 in positions 1, 4, 8, so it corresponds to picking 1, 4 and 8.

The original question is then like counting bit strings of length 50 (since we pick from numbers 1-50) with 9 1's (since we pick a total of 9 numbers), and no two 1's can be adjacent (since that represents picking two consecutive numbers).