There are eight horses.
There is one harness.
The harness can only be arranged one way.
There can only be two horses on each row.
There are four rows.
How many different ways can the horses be arranged?
Diagram:
1 2
3 4
5 6
7 8
Can you solve it any other ways besides the diagram?
If so, how?
Thanks in advance.
:]
2006-11-15 10:27:01 · 4 answers · asked by JJ 1 in Science & Mathematics ➔ Mathematics
A simple 8!
8 choices for the first horse
7 choices for the 2nd horse
6 choices for the 3rd...
...
2 choices for the 7th horse
1 choice for the last horse
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 ways
This assumes that left and right arrangements are different.
2006-11-15 10:30:01 · answer #1 · answered by Puzzling 7 · 0⤊ 0⤋
Your question title says combinations, but the question says how many ways can they be arranged. Are you confused about which is which?
If horse 1 on left, 2 on right in first row is a different ARRANGEMENT from 2 on left, 1 on right, then it's just 8! = 40,320. If something different is meant, you'll have to spell it out.
2006-11-15 18:36:17 · answer #2 · answered by Philo 7 · 0⤊ 0⤋
well, 1 can go in 7 other ways, 2 6, so over all, the problems like this, including the one you gave:
8+7+6+5+4+3+2+1= 36 combinations for the horses
Thats all there is!
2006-11-15 18:34:26 · answer #3 · answered by themastersensei 2 · 0⤊ 1⤋
Doesn't really matter how the horses are arranged as a whole, because there is only 1 possible arrangement, but there are 8!, or 40320, ways to arrange the horses, I'll tell you that.
2006-11-15 18:31:14 · answer #4 · answered by Anonymous · 0⤊ 0⤋