Question 1060867: How many ways can four calculators be chosen for testing from a group of ten?
Answer by addingup(3677)   (Show Source):
You can put this solution on YOUR website! I take it order is not important, it doesn't matter which one you pick first and which one last. Therefore, this is a COMBINATION (vs. a permutation)
:
There are 4! or 4x3x2x1 = 24 different ways the exact same 4 calculators can be arranged. These 24 ways are
ABCD, ABDC, ACBD, ACDB, ADBC, ADCB
BACD, BADC, BCAD, BCDA, BDAC, BDCA
CABD, CADB, CBAD, CBDA, CDAB, CDBA
DABC, DACB, DBAC, DBCA, DCAB, DCBA
:
This is given by:
10C4 = 10!/(6!.4!) = (10*9*8*7)/4! here we simplified, where on top we got to 6! we canceled it with the 6! below the line.
So now we have:
(10*9*8*7)/(4*3*2) = 210 is the number of ways.
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NOTE:
If order was important, then this would be a permutation:
10P4 = 10!/6!
= 10*9*8*7 = 5040 ways
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