If theres a row of 7 desks with 7 people, how many possible combinations of the order of the people can there be?
Just to tell you, the answer is not 49.
I think the answer is 343.
Thanks!!
2006-08-29 12:38:01 · 8 answers · asked by whetherwoman 2 in Science & Mathematics ➔ Mathematics
It is 7! (7x6x5x4x3x2x1) or 5040
Think about it this way, 7 different people could sit in the first seat, but only 6 in the second (because one person is already in the 1st seat), in the 3rd there are only 5 possibilities and so on.
2006-08-29 12:48:57 · answer #1 · answered by firefly 3 · 2⤊ 1⤋
Let's say you are filling the chairs.
The row is empty.
You can put 7 possible people in the first chair.
Then you can put 6 possible people in the second
5 in the third
Then, 4, 3, 2 and finally just one left for the last chair.
For total number of possibilities, multiply the number of possibilities for each chair:
7! (exclamation mark means "7-factorial") = 7*6*5*4*3*2*1
=42*20*6
=840*6
=5040
2006-08-29 23:17:59 · answer #2 · answered by Anonymous · 0⤊ 0⤋
The answer is 7!=7*6*5*4*3*2*1=5040
2006-08-29 19:46:24 · answer #3 · answered by Logic + 1 · 1⤊ 0⤋
If it is combinations(order does not matter), then it is just 1.
If order does matter (this is called permutations), then the answer is 7! = 7*6*5*4*3*2*1 = 840
2006-08-29 19:45:01 · answer #4 · answered by MsMath 7 · 1⤊ 1⤋
Look up "permutations" and "combinations" for a more general way to do these types of problems!
2006-08-29 21:40:13 · answer #5 · answered by tomz17 2 · 0⤊ 0⤋
Hi. !7 (factoral 7). Use your calculater. (7x6x5x4x3x2x1)
2006-08-29 19:43:37 · answer #6 · answered by Cirric 7 · 1⤊ 0⤋
7x6x5x4x3x2x1=5040
2006-08-29 19:44:15 · answer #7 · answered by flutterflie04 5 · 2⤊ 0⤋
whatever
2006-08-29 19:40:50 · answer #8 · answered by Anonymous · 0⤊ 2⤋