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Mathematics 21 Online
OpenStudy (anonymous):

How many 4 letter words can be formed from 'MORADABAD'

OpenStudy (anonymous):

this might help? http://www.solverscrabble.com/solver/

OpenStudy (asdasdasd):

626.

OpenStudy (anonymous):

I can't even think of one. What about DADA lol.

OpenStudy (anonymous):

roam,boar,dorm, etc...i found 18

OpenStudy (anonymous):

can someone please explain this question to me @AkashdeepDeb

OpenStudy (anonymous):

its not scrabble , and no @mathbrz , there r repititions , u cant solve it like that

OpenStudy (anonymous):

@mathbrz erm... actual words?

OpenStudy (anonymous):

@Emz.Yaoi , nope

OpenStudy (akashdeepdeb):

You have to find all possible cases with all the letters included.

OpenStudy (anonymous):

awww well that's no fun XD well then, let the math geniuses handle this then :P

OpenStudy (anonymous):

Oh, so it's more like a math question.

OpenStudy (anonymous):

@Kamizamurai , yesss

OpenStudy (anonymous):

@mathstudent55

OpenStudy (anonymous):

@mathmale

OpenStudy (mathmale):

While I do not know how to obtain the final count of how many such words are in MORADABAD, I can suggest that looking for the number of possible 4-letter permutations could be derived from MORADABAD (9 letters). Of course, not every such permutation would constitute a legitimate, four-letter word. mathbrz has already calculated the number of possible 4-letter permutations, above, and in\[P^9_4 = \frac{9!}{(9-4)!} = 9*8*7*6 = 3024\] but now you'll have to find a way to sort through them and determine which actually are English words.

OpenStudy (anonymous):

Thanks for the help , but real english words are not required , only the different words (with or without meaning that can be formed) Also , ur method doesnt eliminate the repititions

OpenStudy (anonymous):

@mathmale , the answer is 626 , just dunno how to get it

OpenStudy (mathmale):

Wish I were able to help more. Very interesting question, and good discussion.

OpenStudy (anonymous):

hahah , @mathbrz , u cant just eliminate them from the list

OpenStudy (anonymous):

for example the , word 'FEED' no of possible words is 4!/2! = 12 now , according to you , if i make it 'FED; the answer is 3! which is 6 , which is wrong

OpenStudy (anonymous):

look at my example

OpenStudy (anonymous):

OK , guys thanks for the help , Ill try to figure it out later

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