import sys from sys import stdin import heapq import re from itertools import permutations from bisect import bisect_left, bisect_right from collections import Counter, deque from math import factorial, sqrt, gcd from functools import lru_cache, reduce INF = 1 << 60 MOD = 1000000007 sys.setrecursionlimit(10 ** 7) # UnionFind class UnionFind(): def __init__(self, n): self.n = n self.parents = [-1] * n def find(self, x): if self.parents[x] < 0: return x else: self.parents[x] = self.find(self.parents[x]) return self.parents[x] def union(self, x, y): x = self.find(x) y = self.find(y) if x == y: return if self.parents[x] > self.parents[y]: x, y = y, x self.parents[x] += self.parents[y] self.parents[y] = x def size(self, x): return -self.parents[self.find(x)] def same(self, x, y): return self.find(x) == self.find(y) def members(self, x): root = self.find(x) return [i for i in range(self.n) if self.find(i) == root] def roots(self): return [i for i, x in enumerate(self.parents) if x < 0] def group_count(self): return len(self.roots()) def all_group_members(self): return {r: self.members(r) for r in self.roots()} def __str__(self): return '\n'.join('{}: {}'.format(r, self.members(r)) for r in self.roots()) # ダイクストラ def dijkstra_heap(s, edge, n): #始点sから各頂点への最短距離 d = [10**20] * n used = [True] * n #True:未確定 d[s] = 0 used[s] = False edgelist = [] for a,b in edge[s]: heapq.heappush(edgelist,a*(10**6)+b) while len(edgelist): minedge = heapq.heappop(edgelist) #まだ使われてない頂点の中から最小の距離のものを探す if not used[minedge%(10**6)]: continue v = minedge%(10**6) d[v] = minedge//(10**6) used[v] = False for e in edge[v]: if used[e[1]]: heapq.heappush(edgelist,(e[0]+d[v])*(10**6)+e[1]) return d # 素因数分解 def factorization(n): arr = [] temp = n for i in range(2, int(-(-n**0.5//1))+1): if temp%i==0: cnt=0 while temp%i==0: cnt+=1 temp //= i arr.append([i, cnt]) if temp!=1: arr.append([temp, 1]) if arr==[]: arr.append([n, 1]) return arr # 2数の最小公倍数 def lcm(x, y): return (x * y) // gcd(x, y) # リストの要素の最小公倍数 def lcm_list(numbers): return reduce(lcm, numbers, 1) # リストの要素の最大公約数 def gcd_list(numbers): return reduce(gcd, numbers) # limit以下の素数を列挙 def eratosthenes(limit): A = [i for i in range(2, limit+1)] P = [] while True: prime = min(A) if prime > sqrt(limit): break P.append(prime) i = 0 while i < len(A): if A[i] % prime == 0: A.pop(i) continue i += 1 for a in A: P.append(a) return P # 同じものを含む順列 def permutation_with_duplicates(L): if L == []: return [[]] else: ret = [] # set(集合)型で重複を削除、ソート S = sorted(set(L)) for i in S: data = L[:] data.remove(i) for j in permutation_with_duplicates(data): ret.append([i] + j) return ret # ここから書き始める n = int(input()) a = list(map(int, input().split())) b = list(map(int, input().split())) total = factorial(n) ** 2 cnt = 0 for i in permutations(a): for j in permutations(b): acnt = 0 bcnt = 0 for k in range(n): if i[k] > j[k]: acnt += 1 else: bcnt += 1 if acnt > bcnt: cnt += 1 ans = cnt / total print(ans)