2006-07-31 00:46:42 · 8 answers · asked by Amit K 2 in Science & Mathematics ➔ Mathematics
Although you say repetition of digits permitted, we obviously can't have repeated digits, or else two or more digits would be equal, and thus cannot satisfy the inequality a>b>c (if you had said ≥, then they could). So first we find the number of permutations that don't repeat, which is 10*9*8 or 720. Next, we observe that given any three distinct digits, there are 6 ways to order them and only one of those ways satisfies a>b>c. Thus, only one out of every six of these permutations are valid. So the final answer is 720/6 or 120 triplets.
2006-07-31 01:08:03 · answer #1 · answered by Pascal 7 · 4⤊ 0⤋
10 choose 3 = 10!/3!7! = 10x9x8/3x2x1 = 120.
10 choose 3 is the number of different triples that can be selected from10 objects. each such selection {a,b,c} from {0,1,...,9}
corresponds to an ordered triple.
2006-07-31 14:10:03 · answer #2 · answered by bbp8 2 · 0⤊ 0⤋
120
2006-07-31 08:35:39 · answer #3 · answered by DigitalManic 2 · 0⤊ 0⤋
there are 120 triplets possible
starting with 9=36:with8=28:with7=21:with6=15:with5=10:with4=6
with3=3:with2=1
2006-07-31 08:59:24 · answer #4 · answered by raj 7 · 0⤊ 0⤋
288
2006-07-31 08:15:01 · answer #5 · answered by bob h 3 · 0⤊ 0⤋
9 cubed?
or 10 cubed if 0 is a number?
2006-07-31 07:53:33 · answer #6 · answered by ii337 3 · 0⤊ 0⤋
Pardon????
2006-07-31 07:50:06 · answer #7 · answered by ? 5 · 0⤊ 0⤋
say what???????????
2006-07-31 07:50:45 · answer #8 · answered by IrishLassie 4 · 0⤊ 0⤋