2006-09-22 07:50:27 · 5 answers · asked by vinod kumar n 1 in Science & Mathematics ➔ Mathematics
1 is paired with 5 possible others. There's 5 ways to do that.
Then, the lowest of the remaining 4 is paired with one of the 3 other remaining. There's 3 ways to do that.
Then, the remaining 2 are paired. There's 1 way to do that.
All together there are 5*3*1 = 15 ways.
60 is wrong because you must divide by the permutations: for eample, 15 23 46 is the same as 15 64 23, and the method that got 60 as an answer would count this single arrangement as two arrangements
2006-09-22 09:22:52 · answer #1 · answered by vinzklorthos 2 · 0⤊ 0⤋
lets see...hmmm
12 34 56
13 24 56
14 52 36 ...etc
assume any number for face 1 (arbitrary)
then there are 5 choices for its opposite side
once that is done, there are 4 possibilities for the 3rd face
then there are 3 possibilities for its opposite
the remaining two shall go to the remaining faces...no possibilities
hence total possibilities = 5 * 4 * 3 = 60
2006-09-22 14:58:41 · answer #2 · answered by m s 3 · 0⤊ 0⤋
i think its 1 against 6 and 2agnst 5 and 3 agnst 4 such that their sum will always be equal to 7
2006-09-22 14:55:07 · answer #3 · answered by Anonymous · 3⤊ 0⤋
I didn't get the question clearly, is there another dice involved?
I remember the below formula from my probability course i dunno if it applies here
(n-1)!
(6-1)!=5!= 5 * 4 * 3 * 2 * 1 = 120
2006-09-22 15:04:58 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Pryia beat me to the punch. She's right, so rewrite your question!!
2006-09-22 14:59:20 · answer #5 · answered by Nikki 6 · 2⤊ 0⤋