 Hi and how are you all today? I'm Priyanka. The question says, if the different permutation of all letters of the word examination are listed as in a dictionary, how many words are there in this list before the first word starting with E? Let's just start with our solution. Now here, the given word is examination. Right? Now, if the permutation of the letters of this word is listed in dictionary, words before the first word starting with E are all the words starting with letter A, isn't it? So, we fix A as the first letter of the word. Remaining 10 letters T-I-O-N can have certain numbers of permutation. Now, using theorem 4 of permutation, there will be as many words found in letter examination as the number of permutation of 10 letters. Now, here 2 I's are there and 2 N are there. West all are different. So, the number of permutation that is required, number of permutation divided by 2 factorial divided by 2 factorial again. Let us simplify it. Further, we have 10 multiplied by 9 multiplied by 8 by 7, 6, 5, 4, 3, 2 factorial divided by 2 factorial can be written as 2 multiplied by 1. Again, 2 factorial, 2 factorial will get cancelled out, 2 will get cancelled out and the result of this whole will be a left with 10 multiplied by 9 multiplied by 8 multiplied by 7, 6, 5, 2 and 3. That is equal to 9, 0, 7, 2 double 0. This is our required answer. I hope you enjoyed. Take care.