In computer science and discrete mathematics, an [inversion](https://en.wikipedia.org/wiki/Inversion_%28discrete_mathematics%29) is a pair of places in a sequence where the elements in these places are out of their natural order. So, if we use ascending order for a group of numbers, then an inversion is when larger numbers appear before lower number in a sequence.

Check out this example sequence: ```(1, 2, 5, 3, 4, 7, 6)``` and we can see here three inversions
```5``` and ```3```; ```5``` and ```4```; ```7``` and ```6```.

You are given a sequence of numbers and you should count the number of inversions in this sequence.

```Input```: A sequence as a tuple of integers.

```Output```: The inversion number as an integer.

Example:
```python
  count_inversion((1, 2, 5, 3, 4, 7, 6)) == 3
  count_inversion((0, 1, 2, 3)) == 0
```

Difficulty: EASY
Starter code: def count_inversion(sequence):
	

Solutions:
def count_inversion(nums):
	return sum((a > b for (i, a) in enumerate(nums) for b in nums[i + 1:]))


def count_inversion(s):
	return sum((s[i] > s[j] for i in range(len(s) - 1) for j in range(i + 1, len(s))))


def count_inversion(sequence):

	def rec(arr):
		if len(arr) <= 1:
			return (arr, 0)
		(a, ai) = rec(arr[:len(arr) // 2])
		(b, bi) = rec(arr[len(arr) // 2:])
		(res, i, j, count) = ([], 0, 0, ai + bi)
		while i < len(a) and j < len(b):
			if a[i] <= b[j]:
				res.append(a[i])
				i += 1
			else:
				res.append(b[j])
				j += 1
				count += len(a) - i
		return (res + a[i:] + b[j:], count)
	return rec(list(sequence))[1]


from itertools import combinations

def count_inversion(lst):
	return sum((a > b for (a, b) in combinations(lst, 2)))


def count_inversion(sequence):
	return sum((a > b for (i, a) in enumerate(sequence[:-1]) for b in sequence[i + 1:]))


def merge(a, left_index, mid, right_index, aux):
	(i, j) = (left_index, mid + 1)
	inversions = 0
	for k in range(left_index, right_index + 1):
		aux[k] = a[k]
	for k in range(left_index, right_index + 1):
		if i > mid:
			a[k] = aux[j]
			j += 1
		elif j > right_index:
			a[k] = aux[i]
			i += 1
		elif aux[i] <= aux[j]:
			a[k] = aux[i]
			i += 1
		elif aux[j] < aux[i]:
			a[k] = aux[j]
			j += 1
			inversions += mid - i + 1
	return inversions

def count(a, left_index, right_index, aux):
	if right_index <= left_index:
		return 0
	mid = left_index + (right_index - left_index) // 2
	left_inversions = count(a, left_index, mid, aux)
	right_inversions = count(a, mid + 1, right_index, aux)
	return left_inversions + right_inversions + merge(a, left_index, mid, right_index, aux)

def count_inversion(sequence):
	if len(sequence) < 2:
		return 0
	aux = [None for _ in range(len(sequence))]
	sequence_list = list(sequence)
	return count(sequence_list, 0, len(sequence) - 1, aux)


from collections import deque

def count_inversion(nums):
	forward_nums = deque(nums)
	result = 0
	for a in nums:
		forward_nums.popleft()
		for b in forward_nums:
			if a > b:
				result += 1
	return result


def count_inversion(sequence):
	count = 0
	for (i, x) in enumerate(sequence):
		for y in sequence[i + 1:]:
			if x > y:
				count += 1
	return count


count_inversion = lambda sequence: sum([sum([1 if sequence[x] > sequence[x + y] else 0 for y in range(1, len(sequence) - x)]) for x in range(len(sequence))])


def count_inversion(sequence):
	return sum((n > m for (i, n) in enumerate(sequence) for m in sequence[i:]))