If all the letters of the word again

Answer

Verified

Hint: The words in a dictionary are arranged in alphabetical order at each stage. In this question must consider all the words in order starting with A, G, I, N. As in dictionary A will appear at first place so remaining four letters can be filled in how many ways. The number of ways of arranging the remaining 4 letters taking all at a time i.e. the number of words starts with A are $4!$. Similarly, you can find the number of words formed with other letters at first place. We have to find the word at 50th place in the dictionary according to given alphabets.

Complete step-by-step answer:

As the are placed in alphabetical order so the number of words start with letter A is given by:
[ Rightarrow ]Number of words that can be formed with A at first position $ = 4! = 24$.
Rest of the letters can be placed at first place and let us see how many words are formed from them respectively. G letter will come after A letter. So, number of words formed with first letter G is given by
[ Rightarrow ]Number of words that can be formed with G at first position $ = dfrac{{4!}}{{2!}} = dfrac{{4 times 3 times 2 times 1}}{{2 times 1}} = 12$.
Here you can see we divide $4!$ by $2!$ why? The reason behind that is that there are two A letters in the word given to us. So, to avoid repetition of words and the position of the last word appear in a dictionary whose first letter G can be given by the addition of number of words formed by A and I i.e.
$ Rightarrow $The number of words till G letter at starting = $24 + 12 = 36$.
Similarly, the number of words formed with first letter I is given by
 [ Rightarrow ]Number of words that can be formed with I at first position $ = dfrac{{4!}}{{2!}} = dfrac{{4 times 3 times 2 times 1}}{{2 times 1}} = 12$.
Hence, you can find the position of last word start with letter I i.e.
$ Rightarrow $ The number of words till I letter at starting = $24 + 12 + 12 = 48$.
Now you can see the first 48 words appearing in a dictionary are from the words starting with letter A, I, G. After that the next two words i.e. word at 49th and 50th position have starting letter N.
So, look at the first two words that appear in the dictionary starting with the N alphabet are NAAGI and NAAIG in alphabetically order respectively.
Here, the word at 50th position is NAAIG.
Hence the correct option is C.

Note: The common mistake done by students in calculating the number of ways of arranging the letters is repetition when we arrange a similar alphabet the word form is the same. So, we have to divide those repeated words. If a similar alphabet repeats then the number of ways of arranging them is divided with the total ways. Here in this question, the number of letters with A is 2 so divide it by $2!$.

Solution

Starting with the letter A and arranging the other four letters, there are 4! = 24 words. These are the first 24 words. Then, starting with G and arranging A, A, I and N in different ways, there are 4!2!1!1!=12 words, the 37th word starts with I and there are 12 words starting with I. Total goes up to 48 words.

Now, 49th word is NAAGI.

OR

I want to sort out a word from main_word_list which contains all the characters from target_characters

main_word_list=['whets', 'beets', 'pelts', 'vests', 'debts', 'welts', 'seeth', 'jests', 'hefts', 'melts', 'zests', 'depth', 'pests', 'meets', 'teeth', 'lefts', 'stets', 'tests', 'wefts', 'felts', 'wests', 'bests', 'belts']

target_characters=['p', 'h', 'd', 'e']

Here only depth would be the correct answer as it was in the main_word_list as well as it has all the target_characters

[w for w in main_word_list if all(c in w for c in target_characters)]

1

target_characters_set=set(target_characters)
result=[]

for word in main_word_list:
    word_set=set()
    for char in word:
        word_set.add(char)
    if len(target_characters_set-word_set)==0:
        result.append(word)

1

set= ['d'-'a', 'e'-'a', 'h'-'a', 'p'-'a'] # We work with lowercase
mask= 'FFFFFFFFFFFFFFFFFFFFFFFFFF' # Various representations are possible

# Set the flags
for letter in word:
  mask[letter - 'a']= 'T'

# Count the matching flags
if Sum(mask[letter] == 'T' for letter in set) == len(set):
  print(word)

# Reset the flags (alternatively, reset the whole mask)
for letter in word:
  mask[letter - 'a']= 'F'

2

def check(word, list):
    for c in list:
        if c not in word:
            return False
    return True

final = []
for word in main_word_list:
    if check(word,target_characters):
        final.append(word)
print(final)

1