2006-09-20 15:16:39 · 8 answers · asked by beatle 1 in Science & Mathematics ➔ Mathematics
The solution involves combinatorics. Fun word, huh?
The formula for # of combinations (c) possible out of a total of n is: [n*(n-1)*(n-2)*....*(n-c+1)] /[c!]
For your particular problem, you plug your values in like this:
[26*25*24]/[3!] = [15600]/[3*2*1] = 15600/6 = 2600
Thus, there are 2600 different combinations of 3 you can made with 26 teams.
2006-09-20 15:25:10 · answer #1 · answered by Michelle_PhD 2 · 0⤊ 0⤋
nPr if arrangement matters
26!/(26-3)! = 26 x 25 x 24 x 23....x 1 / 23 x 22 x 21 x 21 ... x 1
15600
nCr if arrangement doesn't matter: I think it is this one
2600
2006-09-20 22:25:41 · answer #2 · answered by Bryant 1 · 0⤊ 0⤋
(26!)/(3!(26-3)!)
= (26!)/(3!(23!) Cancel 23!
= (26*25*24)/(3*2)
= 15600/6
=2600 combinations!
2006-09-20 22:58:50 · answer #3 · answered by Chris 5 · 0⤊ 0⤋
26 x 25 x 24 = 15,600
2006-09-20 22:24:24 · answer #4 · answered by Anonymous · 0⤊ 0⤋
26C3= 26!/(3!*23!)= 26*25*24/6
2006-09-20 23:58:31 · answer #5 · answered by tronic_hobbist 2 · 0⤊ 0⤋
"Catastrophe" beat me to the correct answer... so that is now verified... just remember in "COMBINATORICS" you must lay out any progressions/ parameters/ restrictions on the sets to combine... eg, in actual chemistry, "metals" to nonmetals... This was an "open" problem, so it was pretty "straight forward." In real life, valid combinations are not as open... but constraints on variability are part 'n' parcel to the problem when combinatorics come into question.... and this goes back to one of the biggest probems in applied mathematics... that of complexity & "computablity."
PS The "factorial" (!) you use is dependent upon the stated question... read it carefully.
2006-09-20 22:32:19 · answer #6 · answered by cherodman4u 4 · 0⤊ 0⤋
26!/(23!*3!)
=(26*25*24)/6
=26*25*4
=2600
2006-09-20 22:27:06 · answer #7 · answered by banjuja58 4 · 0⤊ 0⤋
nCr
26C3
26!
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(26-3)!3!
2006-09-20 22:22:36 · answer #8 · answered by Catastrophic. 2 · 0⤊ 0⤋