Number of numbers!

Let $a_{1}<a_{2}<a_{3}<a_{4}<a_{5}<a_{6}<a_{7}<a_{8}$ be the $8$ numbers satisfying the conditions above. We take $(b_{1},b_{2},b_{3},b_{4},b_{5},b_{6},b_{7},b_{8})=(a_{1},a_{2}-1,a_{3}-2,a_{4}-3,a_{5}-4,a_{6}-5,a_{7}-6,a_{8}-7)$ . Then $(b_{1},b_{2},b_{3},b_{4},b_{5},b_{6},b_{7},b_{8})$ are $8$ distinct numbers from the first $18$ natural numbers. Conversely, from any $8$ distinct numbers $b_{1}<b_{2}<b_{3}<b_{4}<b_{5}<b_{6}<b_{7}<b_{8}$ , we can reconstruct $(a_{1},a_{2},a_{3},a_{3},a_{4},a_{5},a_{6},a_{7},a_{8})=(b_{1},b_{2}+1,b_{3}+2,b_{4}+3,b_{5}+4,b_{6}+5,b_{7}+6,b_{8}+7)$ to obtain $8$ numbers satisfying the conditions of the problem. Thus we found a one-to-one mapping between the set of $8$ numbers satisfying the given conditions and the set of $18$ distinct numbers from the first $18$ positive integers. Therefore the answer is $\binom{18}{8}=43758$ .

stars and bars technique

Let

• bars | be the 8 numbers to be chosen, total 8 bars.
• stars * be the remaining numbers from the first 25 natural numbers, total 17 stars.

First arrange each bar is separated by 1 star

$|*|*|*|*|*|*|*|$

For the remaining 10 stars,

e.g. if we put 4 at the front and 6 at the end

$****|*|*|*|*|*|*|*|******$

count from left to right, the chosen numbers will be $5,7,9,11,13,15,17,19$

The remaining 10 stars are free to put into 9 urns

$\text{(urn 1)}|*|*|*|*|*|*|*|\text{(urn 9)}$

The number of ways to choose 88 numbers from the first 25 natural numbers such that any two chosen numbers differ by at least 2

= The number of ways to put 10 stars into 9 urns

= ${18\choose 8} = \boxed{43758}$