Possible Combinations (Maths Question)?

Question

How many possible combinations are there if you use three characters and the characters can be letters or numbers?

An example of a combination would be:

2G5 or F69

It can be all letters, all numbers or a mix.

Thanks

Answer 1

There are 26 letters in the alphabet and 10 digits.
Therefore you have 36 possibilities for the first character, 36 for the second for each of the 36 first ones (i.e. 36*36 or 36 squared) and 36 for the third respectively.
Therefore, the answer is 36*36*36 or 36 to the power of 3 (=46656).

Answer 2

A combination means that different orderings of the 3 symbols are treated as being equivalent, so

ABC is the same combination as CBA is the same combination as CAB etc.

Scenario 1
------------------------
Usually in questions like this the rules are that you have a set of symbols and you can use each of them only once. So, on that assumption the set of symbols is A-Z, 0-9 giving a total of 36 symbols.

The rule for working out the number of combinations is:

let S be the total number of symbols you have (36)
let c be the number you are choosing (3)

The number of combinations C = S!/(c! x [S-c]!)

So in this case you have C = 36x35x34/(3x2) = 7140.

One way of seeing how this works is to think about how many choices you have for the first, second and third symbol (36,35, 34 respectively) and then how many orderings are equivalent (6).

If you allow upper and lower case letters in the calculation, then you just work out the above formula but using 62 instead of 36.


Scenario 2
----------------
Suppose you are allowed to use symbols more than once in your selection. Then you would be allowed to have

AAB but this would be just the same as ABA and BAA.

Here you have three possible outcomes:
1) All the symbols are different
2) All the symbols are the same
3) Two of the symbols are the same and the other is different.

Case 1) is just like the one we've already looked at, so we know the answer.

Case 2) ... there are just 36 possibilities and none of them are equivalent

Case 3) ... there are 36*35 possibilities BUT two-thirds of them are duplicates (because AAB equiv. ABA equiv. BAA) so this adds another 36*35/3 = 420 possibilities

So the total for this scenario is

7140+420+36 = 7596 combinations.


(And if you are allowing upper and lower case characters you just work through the logic with 62 instead of36, as before).

Hope that helps.

Answer 3

the question is missing in some details....so we assume:
the alphabets can be lowercase OR uppercase, making a total count of 52 characters

there are 10 numbers, including 0, which can appear anywhere

case 1: assume repetition

each position can be filled in by any of 62 characters; so the total possibilities are: 62*62*62

case 2: assume no repetition

the first position can be filled in by any of 62 characters; the second by 61, the third by 60; so the total possibilities are 62*61*60

case 3: it must begin with an alphabet, no repetition

=52 * 61 * 60

case 4: it must contain at least one alphabet and one number

=52P1 * 10P2 + 52P2*10P1 where mPn means m!/(m-n)!

Answer 4

6

2G5 25G G52 5G2 G25 52G

The rule is x! (x factorial) where x is the number of different characters and factorial means. For example, here we have 3 characters so 3! is 3x2x2 = 6

Or if you mean 0 to 9 and 26 small letters and 26 capital letters x = 62 so the number of possible combinations is 62! which is .... wow Try it on you calculator.

Answer 5

Yes, there is a formula. Because the number in one slot doesn't affects the other (so there could be all possible combinations), the formula is m^n, where m is the number of numbers that could be in that slot and n is the number of slots you have. In your problem you have three slots and 36 numbers (35 and the zero = 36), so the possible combinations are 36^3 = 46656.

Answer 6

there are 9 number you can use and 26 letters so you have 9x26 for the first character.
The second character is the same and so is the same so the answer is 3(9x26) assuming the combinations are not mutualy exclusive, ie you can have the same letter or same number re-occurring.

Anyone agree?

Answer 7

assuming that there are 10 numbers
and 26 letters

36*36*36=46,656 that is if you can
use the elements more than once

36*35*34=42,840 that is if they have to be
all different

i hope that this helps

Answer 8

2G5
25G
52G
5G2
G25
G52

or

2 in first place G in second place 5 in third place
2 in 1st place 5G
5 in 1st place 2G
5 in 1st place G2
G in 1st place 25
G in 1st place 52

Each character will be in a place two times. And, there will be six possible combinations.

Answer 9

let say the combinations are 'abc' and'123'
then the combinations are a12,a23,a13,ab1,ab2,ab3,ac1,ac2,ac3
b12,b23,b13,bc1,bc2,bc3,c12,c23,c13
SO TOTAL 18
THANK U

Answer 10

2gf
2fg
f2g
fg2
g2f
gf2

6 combinatiions