2007-04-04 05:57:16 · 13 answers · asked by Lacie L 1 in Science & Mathematics ➔ Mathematics
5! = 120.
2007-04-04 05:59:53 · answer #1 · answered by Anonymous · 1⤊ 1⤋
2 times
2007-04-04 05:59:39 · answer #2 · answered by JAYSON F 1 · 0⤊ 3⤋
You have 5 spaces to fill and 5 different letters to use to fill them. For the 1st space, you have 5 choices, for the second space you only have 4 choices left, for the 3rd, 3, and so on. Find the total number of permutations by multiplying 5*4*3*2*1, aka 5! (5 factorial).
5! = 120
2007-04-04 06:02:31 · answer #3 · answered by indiana_jones_andthelastcrusade 3 · 1⤊ 0⤋
5! = 5 * 4 * 3 * 2 * 1 = 120 ways
2007-04-04 06:01:08 · answer #4 · answered by I know some math 4 · 0⤊ 0⤋
5 * 4 * 3 * 2 * 1 =120
2007-04-04 06:06:35 · answer #5 · answered by theres_more 2 · 0⤊ 0⤋
5! = 120
2007-04-04 06:08:00 · answer #6 · answered by MAX 1 · 0⤊ 0⤋
5!=120
2007-04-04 05:59:55 · answer #7 · answered by bruinfan 7 · 0⤊ 1⤋
Depends on what you're looking for. Do you need to use all the letters? If not, what's the letter minimum?
If all letters must be used, and if it must be an English word...."house" can not be rearranged.
2007-04-04 06:02:49 · answer #8 · answered by Jenniffer 2 · 0⤊ 0⤋
Are you asking how many words can be made or just random rearrangement?
2007-04-04 06:00:19 · answer #9 · answered by gebobs 6 · 0⤊ 1⤋
If you are talking anagrams:
EH SOU
HE SOU
HOE US
HOUSE
HUE SO
USE HO
USE OH
SUE HO
SUE OH
2007-04-04 06:02:22 · answer #10 · answered by sam 2 · 0⤊ 0⤋