Question 229101: How many combinations of three digits can be made from the numbers 0-9 without repetitions?
Please correct my answer if its wrong.
n=10
r=3
nPr=(n!)/(n-r)!
= (10!)/(10-3)!
= (3,628,800)/ (7)!
= (3,628,800)/5040
= 720
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! How many combinations of three digits can be made from the numbers 0-9 without repetitions?
Please correct my answer if its wrong.
n=10
r=3
nPr=(n!)/(n-r)!
= (10!)/(10-3)!
= (3,628,800)/ (7)!
= (3,628,800)/5040
= 720
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You want combinations, not permutations.
Permutations are arrangements: Example 231 is different than 312
Combinations are groups: Example 231 is the same group as 312
10C3 = 10!/[7!*3!] = 120
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Cheers,
Stan H.
Cheers,
Stan H.
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