how many 4-letter codes can be formed using the lee\tters A B C D E F no letters are repeated fux math!
2006-08-27 18:13:46 · 7 answers · asked by phvxoui 2 in Science & Mathematics ➔ Mathematics
For the first letter you can choose 6; then 5, then 4 then 3 and the position is important.
So the answer is :
6*5*4*3 = 360
2006-08-27 18:18:12 · answer #1 · answered by fred 055 4 · 0⤊ 0⤋
6X5X4X3
If repeating letters it would definately be 6x6x6x6 because 4 spaces(codes) and 6 letters so each space would have 6 letters to choose from.
I am not definately sure of no letters repeating but I am thinking 6x5x4x3. I think this because the first space (code) you have 6 letters to choose from and then the 2nd space you already chose 1 so you have 5 left to choose, and so on for the rest of the spaces (codes)
I hope this is right and good luck!
2006-08-28 01:50:30 · answer #2 · answered by Angelcupcake 3 · 0⤊ 0⤋
The first letter can be chosen in one of 6 ways, The 2'nd letter in one of 5 ways, the 3'rd letter in one of 4 ways, and the last letter in one of 3 ways. So the answer is
6*5*4*3=360 ways.
Doug
2006-08-28 01:20:59 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
128
2006-08-28 01:17:10 · answer #4 · answered by Japan_is_home 5 · 0⤊ 0⤋
128!
2006-08-28 01:16:38 · answer #5 · answered by Bare Azz 2 · 0⤊ 0⤋
6C4=6!/(2!4!)=720/48=15.
2006-08-28 01:28:55 · answer #6 · answered by zee_prime 6 · 0⤊ 0⤋
i find mathguru.com a brilliant site to find all such solutions. check that out!
2006-08-28 03:56:18 · answer #7 · answered by Anonymous · 0⤊ 0⤋