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棋盘左上角到右下角方案数（2）
2022-07-06 12:18:00 【朴小明】
1.从左上角（1,1）到右下角（n，m）的方案数是C（n + m - 2, n - 1）
2.左上角到右下角的方案数之和是 C（n + m - 1, n）- 1
3.第n行的和是C（n + m - 1, n）
第三点的证明，可以看E题的题解Codeforces Global Round 21 A-E - kilo的文章 - 知乎
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