I copied my old post--here you go:
Permutations relate to questions of the type: What is the probability of rolling two fours with six rolls of one fair die. Here is the formula (we all love formulas--they make things such easier)
With the example above set up the problem this way:
1) (n, k)
n= total number of rolls=6
k= number of interest =2 (you want 2 fours)
SO, (6,2)
Pluge 6 and 2 into this equation (you'll love it )
n!
------- X P(favorable)^k X P(unfavorable)^(n-k)
k!(k-n)!
(!) means factorial. Forexample 3! = 3X2X1
(P) means probability
(^) means exponent---"to the"
(X) means multiplied by---its not a variable
Therefore:
6!
---------- X (1/6)^2 X (5/6)^(4)
2!(6-2)!
And that is how you solve for the probability of getting 2(k) fours out of 6👎 rolls.
Now as for a combinitorial question like the one on Top Score which asks about how many cominations of three-groups of roses are possible if you have ten roses. Don't be fooled by different colored roses. To answer this question all you do is use the first expression in the equation above:
n!
------------
k!(n-k)!
This equation relates to the number of possible combinations.
So set it up:
n=10 for ten total roses
k=3 for groups of three
Therefore:
10!
--------- = 120 if I remember correctly.
3!(10-3)!
I know this s a random post but if it helps one person then its all worth it. I hope that when this window maximizes when people open it the equations and signs won't be messed up. I apoligize if they are
