Lets first, we should know what is the RANK of a word?
Rank of a word is the position of that word from the arrangement of letters of that given word in dictionary order. Lets see an example…this can help more in understanding the RANK of a word. Its really interesting to find rank of Word.
Lets took word “RANK”
- How many words, can be created using the letters of word RANK? Ans: The number of characters in word “RANK” is 4. So it will be
4! = 4*3*2*1 = 24.
- Now if we arrange all the 24 words formed by the letters of “RANK” in dictionary order.
- Then at what postion the the word “RANK” will exist?
- That particular position of the word, will be the rank of word “RANK”.
- Take a word which we have to find the rank. I will take “SUNIL”.
- Now arrange the letter of “SUNIL” in alphabatical order.
It will be. “I L N S U“.
- Now take the first letter. Its “I”. Now check, Is the letter “I” is the first letter of “SUNIL”? No.
So the no. of word found can be formed with I will be 4!. I = 4! = 24
- Now go for the second letter. Its “L”. Now check once again is this letter we want in first? No.
So. the no. of words can be formed using “L” at first will be 4!.
L = 4! = 24
- Now go for “N”. Is this we want? No. Write down the number of words can be formed using word. “N” is once again 4!N = 4! = 24
- Now go for “S”. Is this is what we want? Yes. Now cut the letter from the alphabetically ordered word. it will be now. “I L N S U”
Write S and check the word once again in the list.
Is we want SI? No. so the number of words can be formed using SI at first will be. 3! [S] : I -> 3! = 6
- Go for L. is we want SL? No. so it will be 3!. [S] : L -> 3! = 6
- Go for N. is we want SN? No. [S] : N -> 3! = 6
- Go for SU. Is this we want? Yes cut the word U from the list and then it will be “I L N S U“.
Now Go for I. is we want SUI? no. so the no of word can be formed which starts from SUI will be 2! Now [SU] : I -> 2! = 2
- Now go for L. is we want SUL. No. so the no. of words. can be formed. SUL will be 2!. [SU] : L -> 2! = 2
- Now go for N. Is we want SUN? Yes now cut that letter. and this will be “I L N S U“ Is we want SUNI yes. cut that word. Then only one letter will be left.”L”.
- Now go for L. Is we want SUNIL yes. SUNIL words having only 1!.
- [SUN][I][L] = 1! = 1
Now add the whole numbers we get. The sum will be.
24 + 24 + 24 + 6 + 6 + 6 + 2 + 2 + 1 = 95.
So the word SUNIL will be at 95th position if we form the whole words that can be created using the letters of SUNIL arranged in dictionary order.
Lets check the rank of SURYA now.
- A R S U Y//Arranged in Alphabatical Order.
- A = 4! = 24 // Is A We want? No.
- R = 4! = 24// Is R? No.
- [S]:A-> 3! = 6 // Is S? Yes. Cut the word S from that list. And move to next letter. Is SA? No.
- [S]:R-> 3! = 6 // IS SR? No.
- [SU]:A-> 2! =2 //Is SU? Yes. Cut the word U from that list. And move to next letter. IS SUA? No.
- [SUR]:A-> 1! = 1 // Is SUR? Yes. Cut the word A From that list.
- [SURY][A] = 1!= 1 //Is SURY? Yes. Cut Y. Only one letter is now left “A” That will be last. Is “SURYA”? Yes.
Now sum all that. 24 + 24 + 6 + 6 + 2 + 1 + 1 = 64.
Shortcut Method to Find the Rank of WORD
( No Repetition )
Take the Word “SURYA” once again.
- A R S U Y // Alphabatcal Order
- 2 * ( 4 ! ) = 48// Search for S. Remove that word from the list and see how many letter before S? 2 letter.
- 2 * ( 3 ! ) = 12 // Search for U. Remove that letter and count the letters before U now. Its 2.
- 1 * ( 2 ! ) = 2 // Search for R. Remove that letter and count the letters before R. Its 1.
- 1 * ( 1 ! ) = 1// Search for Y. Remove that letter and count the letter before Y. Its 1.
- Add the whole numbers. and add 1 for last letter A.
- Sum: 48 + 12+ 2 + 1 + 1 = 64.
That’s all. So here we see how to find the rank of word if the word has no repetition. In the next post I will tell how to find the rank of a word if the word had repetition like in the case of INDIA. “I” is repeated 2 times.
Edit: Link of Finding Rank in Repetition