Step 1 :

Write all the alphabets in alphabetical order.

Step 2 :

Give a number to each alphabet starting from 0 in alphabetical order.

Step 3 :

In word "MANKIND" first alphabet is "M" which is present in the 4^th position in the Alphabet Box. And after "M", in the word "MANKIND" there are 6 alphabets "ANKIND" which can be arrange in $${{6!} \over {2!}}$$ ways. Here $$2!$$ present because two N presents.

So for alphabet "M" we write $$4.{{6!} \over {2!}}$$

Step 4 :

As in word "MANKIND" we calculated "M", so delete "M" from alphabet box and check remaining alphabets got right numbers or not in alphabet box.

As you can see after deleting "M", 4^th position in the alphabet box is missing so we create a new alphabet box.

Step 5 :

Now next alphabet after "M" in word "MANKIND" is "A" which is present in the 0^th position in the new alphabet box. And after "A" in the word "MANKIND" there are 5 alphabets "NKIND" which can be arrange in $${{5!} \over {2!}}$$ ways.

So, for alphabet "A" we write $$0.{{5!} \over {2!}}$$

Step 6 :

Now in word "MANKIND" we calculated "A", so delete "A" from previous new alphabet box and check remaining alphabets got right numbers or not in alphabet box.

As you can see after deleting "A", 0^th positioin in the alphabet box is missing, so we have to create a new alphabet box.

Step 7 :

Now next alphabet after "A" in word "MANKIND" is "N" which is present in the 3^rd and 4^th position in the alphabet box. We have to choose minimum position for "N" in the alphabet box. So, we choose 3^rd position "N". And after "N" in the word "MANKIND" there are 4 alphabets "KIND" which can be arrange $${{4!} \over {2!}}$$ ways. Here $$2!$$ used because of 2 N's (one "N" is that "N" which we are considering and other "N" which is present in the word "KIND")

So for alphabet "N" we write $$3\, \times \,{{4!} \over {2!}}$$

Step 8 :

Now in word "MANKIND" we calculated "N", so delete 3^rd "N" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting 3^rd "N", 3^rd position is missing in the alphabet box so we have to create a new alphabet box.

Step 9 :

Now next alphabet after "N" in word "MANKIND" is "K" which is present at the 2^nd position in the new alphabet box. And after "K" in the word "MANKIND" there are 3 alphabets "IND" which we can arrange $$3!$$ ways.

So for alphabet "K" we write $$2\times3!$$

Step 10 :

Now, in word "MANKIND" we calculated "K", so delete "K" from alphabet box and check remaining alphabets got right number or not in alphabet box. As you can see after deleting "K", 2^nd position in the alphabet box is missing so we have to create a new alphabet box.

Step 11 :

Now next alphabet after "K" in word "MANKIND" is "I" which is present at 1^st position in the new alphabet box. And after "I" in the word "MANKIND" there are 2 alphabets "ND" which we can arrange in $$2!$$ ways.

So for alphabet "I" we write $$1\times2!$$

Step 12 :

Now in word "MANKIND" we calculated "I", so delete "I" from alphabet box and check remaining alphabets got right numbers or not in alphabet box. As you can see after deleting "I", 1^st position in the alphabet box is missing so we have to create a new alphabet box.

Step 13 :

Now next alphabet after "I" in word "MANKIND" is "N" which is present at 1^st position in the new alphabet box. And after "N" in the word "MANKIND" there is 1 alphabet "D" which we can arrange in $$1!$$ ways.

So for alphabet "N" we write $$1\times1!$$.

Step 14 :

Now in word "MANKIND" we calculated "N", so delete "N" from alphabet box and check remaining alphabets got right numbering or not in alphabet box. As you can see after deleting "N", remaining alphabet "D" got correct numbering so new alphabet box will be same.

Step 15 :

Now next alphabet after "N" in word "MANKIND" is "D" which is present at the 0^th position in the new alphabet box. And after "D" in the word "MANKIND" there is no alphabet so it is the last alphabet.

So for last alphabet we always write $$0!$$

$$\therefore$$ Position of the word "MANKIND" in the dictionary

$$ = 4\,.\,{{6!} \over {2!}} + 0\,.\,{{5!} \over {2!}} + 3\,.\,{{4!} \over {2!}} + 2\,.\,3!\, + 1\,.\,2!\, + \,1\,.\,1!\, + \,0!$$

$$ = 1440 + 0 + 36 + 12 + 2 + 1 + 1$$

$$ = 1492$$