Kick Start 2018-Round H-Problem C. Let Me Count The Ways

Problem

To celebrate the anniversary of Googleland, N couples are going to go for a boat ride in a rowboat. The rowboat is very long, but it is only one person wide, so the people will sit in a line from front to back.

However, during a rehearsal of the voyage, the boat did not move! After investigating, the organizers found that some newlywed couples were not rowing, but writing love poems for each other the whole time. Specifically, there are M pairs of newlywed couples. If the two members of a newlywed couple are sitting next to each other, they will be so busy writing poems that they will not row.

Now the organizers have come to you, the smartest person in Googleland, to ask, how many possible ways are there to arrange all 2N people on the rowboat, such that for each of the M newlywed couples, the two members are not sitting next to each other? Two ways are different if there is some position in the boat at which the two ways use different people. Notice that for the purpose of counting the number of ways, the two members of a couple are not considered to be interchangeable. Since the number can be very large, the organizers only want to know the value of the answer modulo 1000000007(109+7).
Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of one line with two integers N and M as described above.
Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the number of possible arrangements, modulo 1000000007(109+7).
Limits

1 ≤ T ≤ 100.
Small dataset

1 ≤ M ≤ N ≤ 100.
Large dataset

1 ≤ M ≤ N ≤ 100000.
Sample

Input
5
2 1
2 2
3 1
3 2
10 5

Output
Case #1: 12
Case #2: 8
Case #3: 480
Case #4: 336
Case #5: 560963525

In Sample Case #1, there are 2 couples. To make the description simpler, we use the characters A and a to represent the newlywed couple, and B and b to represent the other couple. Per the rules of the problem, A and a cannot be adjacent. There are 12 ways to arrange the four people:
ABab ABba AbaB AbBa
aBAb aBbA abAB abBA
BAba BabA bABa baBA

In Sample Case #2, both two couples are newlywed couples, so A and a cannot be adjacent, and B and b cannot be adjacent. They can be arranged in the following 8 ways:
ABab AbaB aBAb abAB
BAba BabA bABa baBA

import math
def c(m,n):
    a = math.factorial(m)//math.factorial(n)
    return a//math.factorial(m-n)

t = int(input())
for case in range(t):
    [n,m] = input().split(' ')
    n = int(n)
    m = int(m)
    l = 2*n
    count = 0
    flag = -1
    for i in range(m+1):
        flag = -flag
        a = c(m,i)
        b = math.factorial(l-i)
        d = 2**(i)
        count += a*b*d*flag
        # print(a,b,d,count,flag)
        # if count<0:
        #     print(count)
        # (_,count) = divmod(count,1000000007)
    (_, count) = divmod(count, 1000000007)
    print("Case #",end='')
    print(case+1,end='')
    print(": ",end='')
    print(count)

这道题花了两个小时。。因为真心有点麻烦，只通过了small的没有通过large的。
理解题意不难，输入是两个数字，前一个数字代表有多少对couple，比如2代表是AaBb两对夫妇，然后第二个数字代表有多少对闹矛盾的夫妇。闹矛盾的夫妇是不可以坐在一起的不然他们只会打架就不来划船了。然后让你算有多少组方法。
这道题放在高中就是一个难一点的排列组合问题嘛。。
搜了一下，这种题目是有公式的，代公式嘛，我寻思这时间复杂度也不高啊，难道是计算很耗时吗。（那也只能是计算很耗时了）。总之大数据集没有跑出来。