4个可重复取值位置的分组元素数量计算公式求解

Let's work through these combinatorial counting problems for your four-position (m, n, l, p) setup where each position can take values from 0 to N-1 (with repeats allowed). I'll break down each case clearly:

This is the simplest case—we just need to pick a single value that all four positions share. Since there are N possible values (0 to N-1), the total number of such combinations is exactly N.

For example, if N=3, the valid combinations are (0,0,0,0), (1,1,1,1), (2,2,2,2) — that's 3 total, which matches N.

This covers the four subcases: (m=n=l)≠p, (m=n=p)≠l, (m=l=p)≠n, (n=l=p)≠m. Let's break down the logic:

• First, choose the value that the three matching positions will take: there are N options.
• Next, choose a value for the fourth position that's different from the first: that gives N-1 options.
• Finally, there are 4 distinct ways to choose which position is the "odd one out" (p, l, n, or m).

Multiply these together, and the total number of combinations is 4 * N * (N-1).

Using N=3 as an example: 432=24. If you list them out, you'll find 6 combinations for each "odd position" (e.g., 3 values for the trio, 2 for p: 3*2=6, times 4 positions gives 24 total).

Here we're looking for combinations where m and n are the same, l and p are the same, but the two pairs have different values. Here's how to calculate it:

• Pick a value for the first pair (m,n): N choices.
• Pick a value for the second pair (l,p) that's not equal to the first pair: N-1 choices.

Since the pairs are fixed to specific positions (m/n vs l/p), we don't have to worry about double-counting (e.g., (0,0,1,1) is a distinct combination from (1,1,0,0)). So the total number of combinations is N * (N-1).

For N=3, that's 3*2=6 valid combinations: (0,0,1,1), (0,0,2,2), (1,1,0,0), (1,1,2,2), (2,2,0,0), (2,2,1,1).

内容的提问来源于stack exchange，提问作者Felipe Augusto de Figueiredo