In this

**Hackerrank Iterables and Iterators**problem we have**Input Format**

The input consists of three lines. The first line contains the integer, denoting the length of the list. The next line consists of space-separated lowercase English letters, denoting the elements of the list.

The third and the last line of input contains the integer, denoting the number of indices to be selected.

**Output Format**

Output a single line consisting of the probability that at least one of the indices selected contains the letter:”.

**Note:**The answer must be correct up to 3 decimal places.

## Problem solution in Python 2 programming.

import math def nCr(n,r): f = math.factorial return f(n) / f(r) / f(n-r) N = input() letters = raw_input().strip().split() K = input() N_a = letters.count('a') if N_a == 0: print '0' elif N - N_a < K: print '1' else: num = nCr(N - N_a, K) denum = nCr(N, K) print (denum - num) * 1.0 / denum

## Problem solution in Python 3 programming.

# Enter your code here. Read input from STDIN. Print output to STDOUT from itertools import combinations N = int(input()) L = input().split() K = int(input()) C = list(combinations(L, K)) F = filter(lambda c: 'a' in c, C) print("{0:.3}".format(len(list(F))/len(C)))

### Problem solution in pypy programming.

from __future__ import division import itertools n = int(raw_input()) alphas = raw_input().split() k = int(raw_input()) checkSymbol = 'a' combinations = list(itertools.combinations(alphas,k)) filtered = [cb for cb in combinations if checkSymbol in cb] print len(filtered)/ len(combinations)

### Problem solution in pypy3 programming.

from itertools import combinations # Enter your code here. Read input from STDIN. Print output to STDOUT n = int(input()) lst = [i for i in input().split()] k = int(input()) a_indxs = {i for i in range(1, n + 1) if lst[i - 1] == 'a'} a_num = 0 comb_len = 0 for comb in combinations(range(1, n + 1), k): comb_len += 1 if set(comb) & a_indxs: a_num += 1 print(a_num / comb_len)