MATHFAGS PLEASE HELP
If I have 48 colors and can make combinations of 2 using any of the 48 colors, how many total combinations can I make?
MATHFAGS PLEASE HELP
If I have 48 colors and can make combinations of 2 using any of the 48 colors, how many total combinations can I make?
1098
>inb4 "BRAINLET"
There's gotta be at least 48 tho, rite?
unless you are rich and will be paying for anwers, don't ask on here
you can make dubs
2256
If you have to use two different colors, the answer is 48 * 47 combinations. If you can repeat a color, the answer is 48 * 48 combinations.
a few
Two tries and not one of those is under 48...
I think user is onto something...
n!/(n-k)!
48! / (48-2)!
2256
What even is the formula to solving this problem?
If you repeat a color you just end up with the same color...
What do those exclamation points mean in math? Whenever I see those I always read NOT
factorial
Fuck off, samefag. We're hard at work here
Which means you multiply by every number leading up to it.
Example:
4! = 4*3*2*1 = 24
This isent programming mate, they mean like factorial.
5 ! = 5 × 4 × 3 × 2 × 1 = 120.
It means factorial.
1! = 1
2! = 2*1 =2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
And so on
0! = 1
If you can mix two of the same colour:
48^2 / 2 = 1152
If mixing the same colour twice isn't allowed:
48 * 47 / 2 = 1128
You have to divide by 2 because if you mix green then yellow, or yellow then green, you get the same result.
This guy gets it
1128
For example look at this hexagon, the total number of combinations for all corners would be 5 + 4 + 3 + 2 + 1 = 15 (15 lines total).
For a 48 sided polygon it would be 47 + 46 + 45 + ... + 2 + 1 = 1128
True,the forumalas used who gets 2256 combinations care about order, in this case order does not matter.
Formula for this is: n!/((n-k)!*k!)
n=48, k=2
((or just nCr(48, 2) for the rest of us humans))