2006-07-03 09:44:58 · 14 answers · asked by Tony / Rose S 2 in Science & Mathematics ➔ Mathematics
I think you are asking how many three-number combinations can be drawn from a set of thirty numbers. If that is right ...
Drawing three out of thirty would be calculated as:
(30!)/(27!x3!)
= (30x29x28)/(3x2x1)
= 4060
2006-07-03 09:57:35 · answer #1 · answered by tdw 4 · 2⤊ 0⤋
3
2006-07-03 16:53:40 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Your first choice is 1 out of 30.
The next one is 1 out of 29 and the last one is 1 out of 28.
So you have 30 x 29 x 28 possibilities, being 24360.
2006-07-03 16:59:08 · answer #3 · answered by Thermo 6 · 0⤊ 0⤋
David is close, but he is describing permutations. You need to divide by 3! in order to compensate for multiple counting.
C(30,3) = 30*29*28/(1*2*3) = 4060.
2006-07-03 16:51:18 · answer #4 · answered by Ѕємι~Мαđ ŠçїєŋŧιѕТ 6 · 0⤊ 0⤋
30*29*28= 24 360
24,360 is your answer
2006-07-03 16:48:15 · answer #5 · answered by Nieds 2 · 0⤊ 0⤋
What are the 30 numbers? 1-30, then uhh, i need to think about it, lol.
2006-07-03 16:48:12 · answer #6 · answered by Shorty 4 · 0⤊ 0⤋
depends on what 30 numbers
2006-07-03 16:47:50 · answer #7 · answered by Dustfinger the master of fire 3 · 0⤊ 0⤋
assuming that numbers can't be repeated
nPr = (n!)/((n - r)!)
30P3 = (30!)/((30 - 3)!)
30P3 = (30!)/(27!)
30P3 = 30 * 29 * 28
30P3 = 24360
2006-07-03 21:37:52 · answer #8 · answered by Sherman81 6 · 0⤊ 0⤋
if they can't repeat = 24360
if they can repeat = 27000
Not Repeatable:
# of permutations of "n" objects taken "r" at a time is:
P(n,r)=(n!)/((n-r)!)
P(30,3)=(30!)/((30-3)!)
P(30,3)=(30!)/(27!)
P(30,3)=24360
Repeatable:
30!
27000
2006-07-03 16:52:39 · answer #9 · answered by Anonymous · 0⤊ 0⤋
30x30x30 if numbers can repete or 30x29x28 for unique combinations.
2006-07-03 16:47:49 · answer #10 · answered by Anonymous · 0⤊ 0⤋