Question 463077: pls help... answer ASAP :
a three digit code is made up of three diff. digts from the set (2,4,6,8,) . how many three digit codes can be formed if no digit can be repeated?... pls explain and show the solution......
Found 2 solutions by Theo, richard1234:
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! the set of digits is 2,4,6,8
you are taking a set of 3 out of a set of 4.
if order is not important, then you are taking a combination of 3 out of 4.
the formula for that is n! / (x! * (n-x)!)( which equals 4! / (1! * 3!) which becomes 4.
if order is important, than you are taking a permutation of 3 out of 4.
the formula for that is n! / (n-x)! which equals 4! / 1! which equals 24.
the number of combinations is the number of sets that have at least one element different from any other set.
your possible combinations are:
246
248
268
468
within each of these sets, you can make 6 arrangements by changing the order of each element in the set.
example:
246 possible arrangements are:
246
264
426
462
624
642
similarly you can do the same with with the other 3 sets.
4 sets with 6 arrangements within each set gets you the 24 permutations.
if the digits could be repeated, the formula would be changed as follows:
you would have 4^3 possible arrangments which would be equal to 64 possible arrangements.
the first number could be any of 2,4,6,8
the second number could be any of 2,4,6,8
the third number could be any of 2,4,6,8
4 possible arrangments for each digit gets you the 4^3.
examples would be:
222
223
233
333
334
344
444
445
455
555
etc,
showing you 64 possible arrangements is tedious.
assume you had 3 numbers and you chose 2 out of 3.
if the numbers could be repeated than you could have 3^2 = 9 possible arrangements
let the numbers be 1, 2, 3, and the possible arrangements would be:
11
12
13
21
22
23
31
32
33
this assumes the same numbers can be used more than once.
if the same numbers could not be used more than once, then the formulas shown above would apply.
the number of combinations would be n! / (x! * (n-x)!) which would be equal to 3! / (2! * 1!) which would be equal to 6 / 2 which would be equal to 3.
those combinations would be:
12
13
23
the number of permutations would be n! / (n-x)! which would be equal to 3! / 1! which would be equal to 6.
those permutations would be:
12
21
13
31
23
31
because the numbers were smaller, i could show you the details of how this works out.
in your problem, the answer is:
number of combinations equals 4! / (3! * 1!) = 4
number of permutations equals 4! / (1!) = 4 * 3 * 2 = 24
Answer by richard1234(7193)  (Show Source):
You can put this solution on YOUR website! There are four choices for the first digit, three for the second digit, and two for the third digit (this is so that no digits repeat). The number of ways is just 4P3 = 24.
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