1. ## Counting Problem

I simply don't know how to start.

In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.

I know that if the "difference between 2" wasn't there it would be C(100,10). But with the difference there how the heck would you set it up?

thanks

2. Originally Posted by bfpri
In how many ways can a set of 10 positive integers less than 101 (100 integers) be selected such that the difference between any 2 numbers in the set is at least 2.
The idea is simple. We can say include $13~\&~15$ among the ten because $|13-15|\ge 2$ but not $15~\&~16$.

So how many ways can we pick ten of those numbers so no two are consecutive?
How many ways can you arrange a string of ten 1’s and ninety 0’s so no two 1’s are next each other?

3. Originally Posted by Plato
The idea is simple. We can say include $13~\&~15$ among the ten because $|13-15|\ge 2$ but not $15~\&~16$.

So how many ways can we pick ten of those numbers so no two are consecutive?
How many ways can you arrange a string of ten 1’s and ninety 0’s so no two 1’s are next each other?
That would be 90!/10!*80! if you split it into 10 "10"s and 90 "1"s.

4. Actually it would be $\dbinom{91}{10}$.
The ninety zeros create ninety-one places to place the ones.

5. Hmm, still don't quite understand why it's 91.

6. Simply using, $1111000000$, four ones and six zeros.
How many ways to arrange that string so no two ones are together?
Look at $\_\_0\_ \_0\_ \_0\_ \_0\_ \_0\_ \_0\_\_$.
Do you see there are seven places to put the four ones: $\dbinom{7}{4}$.
So ninety zeros create ninety one places: $\dbinom{91}{10}$.

7. Got it...thanks!