You are going to be given a word. Your job will be to make sure that each character in that word has the exact same number of occurrences. You will return `true` if it is valid, or `false` if it is not.

For example:

`"abcabc"` is a valid word because `'a'` appears twice, `'b'` appears twice, and`'c'` appears twice.  
`"abcabcd"` is **NOT** a valid word because `'a'` appears twice, `'b'` appears twice, `'c'` appears twice, but `'d'` only appears once!  
`"123abc!"` is a valid word because all of the characters only appear once in the word.

For this kata, capitals are considered the same as lowercase letters. Therefore: `'A' == 'a'` .

#Input 
A string (no spaces) containing `[a-z],[A-Z],[0-9]` and common symbols. The length will be `0 < string < 100`.

#Output
`true` if the word is a valid word, or `false` if the word is not valid.

Source: codewars | Difficulty: EASY
Tags: String algorithms

Starter code: def validate_word(word):
	

Solutions:
from collections import Counter

def validate_word(word):
	return len(set(Counter(word.lower()).values())) == 1


def validate_word(word):
	word = word.lower()
	return len(set((word.count(c) for c in word))) == 1


from collections import Counter
validate_word = lambda word: len(set(Counter(word.lower()).values())) == 1


validate_word = lambda w: len(set((w.lower().count(e) for e in set(w.lower())))) == 1


def validate_word(word):
	word = word.lower()
	c_count = word.count(word[0])
	for c in word:
		if word.count(c) != c_count:
			return False
	return True


def validate_word(word):
	word = word.lower()
	return all((word.count(i) == word.count(word[0]) for i in word))


def validate_word(word):
	word = word.lower()
	arr = list(set(word))
	for i in range(1, len(arr)):
		if word.count(arr[i]) != word.count(arr[i - 1]):
			return False
	return True