SOLUTION: We are working on combinations and permutaions which involve factorials most of the time. How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters c

Question 35133: We are working on combinations and permutaions which involve factorials most of the time.
How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters can repeat themselves and the letter do not have to form a word. How do you solve this?
Found 2 solutions by venugopalramana, Nate:
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
We are working on combinations and permutaions which involve factorials most of the time.
How many 4 letter combinations can be made from the letters f,g,h,i,j & k. No letters can repeat themselves and the letter do not have to form a word. How do you solve this?
COUNT TOTAL NUMBER OF LETTERS=6
TO MAKE 4 LETTER COMBINATIONS YOU SAID THAT IS NOT PROPER I THINK..THAT MEANS A TEAM OF 4 PLAYERS FROM 6 PERSONS.THAT MEANS ABCD..OR...BCDA...OR...CDBA..ETC...ARE ALL SAME TEAM.THE ARRANGEMENT IS NOT IMPORATANT.THEN ONLY IT IS CALLED COMBINATION.THEN THE ANSWER IS 6C4..THE FOMULA FOR NCR = N!/{(R!)*(N-R)!}
SO WE GET 6!/2!*(4-2)!=6!/2!*4!=6*5*4*3*2*1/2*1*4*3*2*1=15...
BUT I THINK YOU WANT ARRANGEMENTS .THEN ITS ANSWER IS 6P4..THE FORMULA FOR
NPR=N!/(N-R)!
6P4=6!/(6-4)!=6!/2!=360
Answer by Nate(3500) (Show Source): You can put this solution on YOUR website!
f,g,h,i,j & k
When you say no letters can repeat themselves, you do not have to worry there are no two of the same letters!
P(6,4)=6!/(2!)=360
Just as venugopalramana said, that is the many ways, but schools usually put permutations as P(n,r) notation ....
P(n,r) is the same as nPr
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