25. Find the number of ways of arranging 13 identical blue balls and 5 identical red balls on a straight line such that between any 2 red balls there is at least 1 blue ball.
25. Answer: 2002
First we place the 5 red balls on a straight line and then place 1 blue ball between 2 adjacent red balls. With this arrangement fixed, the condition of the question is satisfied. We are now left with 9 blue balls. We can place the remaining 9 blue balls into the spaces before, after or in between the 5 red balls. The number of ways that this can be done is the answer to the question. Including the two ends, there are $4+2=6$ spaces into which these 9 blue balls can be placed. The number of ways of distributing the 9 blue balls into the 6 spaces is
$$
\binom{9+6-1}{9}=\binom{14}{9}=\binom{14}{5}=2002
$$