关于先放置减号的思路求解无相邻减号排列问题的技术问询

Hey there! Awesome question—you’re already on the right track with this alternate approach, let’s walk through it to tie up that loose end.

First, let’s recap your setup to make sure we’re aligned:

• We want to count the number of valid arrangements of p plus signs (+) and q minus signs (-) where no two - are adjacent.
• Your starting move: Lay down all q - in a row first: - - - ... - (q total).
• To guarantee no - are adjacent, you insert one + between each pair of -—that uses up q-1 +, leaving you with p - (q-1) = p - q + 1 + left to place freely.

Here’s how to finish the count:

1. Identify the available slots for the remaining +: When you have the base structure - + - + - ... -, there are actually q+1 possible slots to add more +:

   • The slot before the first -
   • The slot between each - and the already-placed + (you can stack more + here)
   • The slot after the last -
     So visually, it looks like: [slot1] - [slot2] + - [slot3] + ... + - [slotq+1]

2. Count the ways to distribute the remaining +: This is a classic "stars and bars" problem. We need to count how many ways to place n = p - q + 1 identical items (the leftover +) into m = q+1 distinct slots, where slots can have 0 or more items. The formula for this is:
   C(n + m - 1, m - 1)
   Where C(a, b) is the combination function "a choose b".

3. Plug in the numbers: Substitute n = p - q + 1 and m = q+1 into the formula:

   C( (p - q + 1) + (q + 1) - 1, (q + 1) - 1 ) = C(p + 1, q)
   

   That’s exactly the same result as the standard "place + first, then choose slots for -" method!

The key here is recognizing that this approach is just the reverse of the standard method—instead of building around the +, you’re building around the - and using stars and bars to account for the flexible placement of extra +. Every valid arrangement can be generated this way, so the count is spot-on.

备注：内容来源于stack exchange，提问作者Rui