# 求组合数
"""
输入: n (求组合数的次数)
输入两个数,即C(a,b)中的a,b
输出: C(a,b)的结果
方法: 建立矩阵,通过公式计算
递推公式,帕斯卡三角形是一个三角形的数字阵列
每一行表示组合数,每一项等于其正上方和左上方两项之和。
c[0][0] = 1
c[1][0] = 1, c[1][1] = 1
c[2][0] = 1, c[2][1] = 2, c[2][2] = 1
c[3][0] = 1, c[3][1] = 3, c[3][2] = 3, c[3][3] = 1
c[4][0] = 1, c[4][1] = 4, c[4][2] = 6, c[4][3] = 4, c[4][4] = 1
"""
N = 2010 # 定义足够大的数用于创建二维数组
mod = 1e9+7 # 用于对结果进行取模运算,以防止整数溢出
# 建立一个N*N的全零二维数组
c = [[0 for i in range(N)] for j in range(N)]
def get():
global c
for i in range(N):
for j in range(i+1):
if j == 0: # 处理边界情况,C(i,0)=1
c[i][j] = 1
else:
c[i][j] = (c[i-1][j] + c[i-1][j-1])%mod
n = int(input())
get()
for k in range(n):
a,b = map(int,input().split())
print(c[a][b])
参考
https://blog.youkuaiyun.com/m0_54689021/article/details/125573187