How many combinations are there of 12 symbols using 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, A, B, C, D, E, F.
2007-07-29 20:18:07 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
The combination function, read as "x choose y", is
Com[x, y] = x! / (y! * (x-y)!)
where ! indicates the factorial function.
16 objects, choosing 12 of them gives us
Com[16, 12] = 1820
So, there are 1820 unique ways to select 12 objects out of 16.
2007-07-29 20:22:29 · answer #1 · answered by lithiumdeuteride 7 · 1⤊ 1⤋
There are two different ways,
1. If in every combination no letter or number could be repeated (for example 111222333444 is not permitted), then for the first letter or number, you have 16 choices. For the second one, you have 15 choices and it goes on. So for the 12th one we have 5 choices. Therefore, we can make 16x15x14x...x5 combinations. you can calculate it by 16!/(16-12)! = 16!/4!
2. If repetition is permitted, then for all letters, from the first to the 12th, we have 16 choices. Thus, we can have 16^12 combinations.
2007-07-30 03:38:21 · answer #2 · answered by Arash 3 · 0⤊ 0⤋
It's 12! = 12x11x10x9x8x7x6x5x4x3x2x1
2007-07-30 04:12:52 · answer #3 · answered by vlee1225 6 · 0⤊ 1⤋
the first answers right if you want the permutation its goin to be
479,001,600*13*14*15*16=20922789888000
ps. did tht in the head......might wanna check
2007-07-30 03:34:49 · answer #4 · answered by Nishant P 4 · 0⤊ 1⤋
16C12= 16C4 =16x15x14x13/4x3x2x1 =1820
2007-07-31 09:20:11 · answer #5 · answered by mramahmedmram 3 · 0⤊ 1⤋