If all words formed using only letters of the word "RANK" exactly once are arranged alphabetically. Find the position of the word RANK.
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Shouldn't it be 60? I get it that you just counted the 4 words. But if I were to follow the question "If all words formed using the letters of RANK....", then there must be 1 letter words, 2 letter words and 3 letter words. And alphabetically they should be on the upper side. So first counting 1 letter words there are 4, similarly 2 and 3 letter words are 12 and 24 respectively. Then comes 4 letter words. So answer should be 60. Did I do make any mistakes?
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@Imrul Khan You're absolutely right, ubt the problem has been rephrased now :)
Nice observation man.
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Ah, don't mention it. It was a cool one you uploaded
The total number of words that can be formed using R, A, N, K are 4! i.e. 24.
Now the words that can be formed after RANK are RKAN, RKNA, RNAK, RNKA.
So RANK's "rank" is 24 -4 = 20
What about RAKN? If you hold the first letter, e.g. R, then there are 3 x 2 x 1 = 6 other possibilities to form the letters. RANK therefor has in total 6 permutations (RANK included). This applies to ALL of the four letters. Which means that 24 should be divided by 4, to know that there are 6 groups starting with the same letter.
4x3x2x1 = 24 words in total, of which 6 start with an A, 6 start with a k and 6 with an N. There are also 6 words starting with an R. A is the first letter in the alphabet, so RAKN is the 19th word. RANK is therefore the 20th word.
if we rank the letters of the word RANK alphabetically: A = first, K = second, N = third and R = fourth.
We can form 4 ! = 2 4 words.
no. of words that starts with A = 1 ( 3 ) ( 2 ) ( 1 ) = 6
no. of words that starts with K = 1 ( 3 ) ( 2 ) ( 1 ) = 6
no. of words that starts with N = 1 ( 3 ) ( 2 ) ( 1 ) = 6
we are finished with the 1 8 positions
next word is
R A K N ---> 1 9 t h p o s i t i o n
then
R A N K ---> 2 0 t h p o s i t i o n
c++ code
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