A phone company wants to create some new area codes. Each new area code will have 3 numbers. Using the digits 1, 5 and 7 only once in each code, how many new codes can the phone company create?
Tell me how to figure this out.
2007-05-31 10:24:23 · 7 answers · asked by steve s 1 in Education & Reference ➔ Homework Help
Ok, since you only have 3 numbers, its not that hard to write out:
157, 175, 517, 571, 751, 715
So you have six codes you can make. The trick is it said you can only use each number once in each code, so there's no such code you can make that has one of the numbers repeating, like 551, or 777....that's when it gets a little more complicated.
2007-05-31 10:32:12 · answer #1 · answered by beatle14 2 · 0⤊ 0⤋
Just write it out. Here are the possibilities:
157
175
517
571
715
751
You could go all statistical and use formulas and such to figure it out, but they are much to complex for such a simple problem. There are six different codes that can be made with 3 different numbers, using each one only once. Hope that helped!
2007-05-31 10:37:35 · answer #2 · answered by anchorsaway 2 · 0⤊ 0⤋
Well simple you just say well the orders could go numerically like 157 then you could go the opposite with 751 and then you could have 1 in the front again and go 175 and then you could have 7 in the front again and go 715 and then you could finnally have 5 in the front and have 517 and 571 with a total of 6 options
2007-05-31 10:35:02 · answer #3 · answered by Ema 2 · 0⤊ 0⤋
1 5 7
1 7 5
5 1 7
5 7 1
7 5 1
7 1 5
Six different codes
3 (3 choices for the first number)
x 2 (times 2 choices for the second number)
x 1 (times 1 choice left for the third number)
= 6
-1 ----5 ---7
-/.\...../.\...../.\
5.7..1.7..1.5
/...\..../..\..../...\
7..5..7.1..5...1
Tree diagram,
showing 3 choices x 2 choices x 1 choice = 6 combinations
2007-05-31 10:30:55 · answer #4 · answered by Nghiem E 4 · 0⤊ 0⤋
becuase its 3 numbers and only 3 digits, its a factoral (written like 3!)
to solve, just multiply 1x2x3 (6) theres ur answer!
factorals are just all the numbers up to that one like 6! would be 1x2x3x4x5x6
good luck!!!
2007-05-31 10:44:31 · answer #5 · answered by Anonymous · 0⤊ 0⤋
157, 175
517, 571
715, 751
you can only make two combinations with the same number begining both so 6
2007-05-31 10:32:40 · answer #6 · answered by g_fredo_99 5 · 0⤊ 0⤋
it's 3! (three factorial) which is
3!=3*2*1 which is 6
2007-05-31 10:31:51 · answer #7 · answered by ladie_lada 2 · 0⤊ 0⤋