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Mathematics 18 Online
OpenStudy (anonymous):

please help and explain this

OpenStudy (anonymous):

http://prntscr.com/1aevbn

OpenStudy (anonymous):

8!/(3! * 3! *2!)

OpenStudy (anonymous):

could you explain in detail

OpenStudy (anonymous):

OpenStudy (anonymous):

thanks @julian25 could you explain me one more question of similar type?

OpenStudy (anonymous):

post it

OpenStudy (anonymous):

How many words are possible using the letters in the word *APPLE*. If the letters may not be repeated.

OpenStudy (anonymous):

for the fisrt position of the word u have 5 posibles leters to chose from for the second position of the word u could chose 4 leters an so on so u get 5! =5*4*3*2*1

OpenStudy (anonymous):

no, but there are two P's and they should not be repeated, if i do it your way i won't be applying the condition above.

OpenStudy (anonymous):

u r right 5! is if u could repeating leters 5!/2! is if the leter may not be repeated

OpenStudy (anonymous):

so the logic behind it is same as for the first question? here also we are using the property of distinguishable permutation?

OpenStudy (anonymous):

@julian25 ?

OpenStudy (anonymous):

yes it is the same logic

OpenStudy (anonymous):

Thank you :D

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