Help with probability problems

[sneaky666]

Homework Statement

1. Suppose we roll 10 fair 6-sided dice. What is the probability that there are exactly two 2's showing?
2. Suppose we are dealt five cards from a standard 52-card deck. What is the probability that
a) we get all 4 aces and the king of spades
b) all 5 are spades
c) we get no pairs (all are different values)
d) a full house (3 of a kind and 2 of a kind)

Homework Equations

This one is correct:
there are 2 pots
in pot1 there is 5 red balls and 7 blue balls
in pot2 there is 6 red balls and 12 blue balls
3 balls are chosen randomely from each pot
chances of all 6 balls to be same color = P(A)
chances of all 6 balls to be red = P(B)
chances of all 6 balls to be blue = P(C)

P(A) = P(B or C)
=P(B) + P(C) -0
P(B)=|B|/|S| = |B|/(12choose3)(18choose3) =
(5choose3)(7choose0)(6choose3)(12choose0)/(12choose3)(18choose3) = 5/4488
P(C)=|C|/|S| = |C|/(12choose3)(18choose3) =
(5choose0)(7choose3)(6choose0)(12choose3)/(12choose3)(18choose3) = 35/816
P(A) = 5/4488 + 35/816 - 0 = 395/8976

The Attempt at a Solution

1.
number of outcomes = 6^10 = 60466176
10!/2!8! = 45
so i get
45/60466176

2.
number of outcomes = 52x51x50x49x48 / 5x4x3x2x1 = 2598960
a) Here i have two different methods, i don't know if both are wrong or one is right...
method 1
(4choose1 * 4choose4 ) / 2598960 = 1/649740
method 2
( (13choose1) * (13choose1) * (13choose1) * (13choose1) ) / 2598960 = ~0.066
b)
(13choose5)/2598960 = ~4.95x10^(-4)
c)
( (4choose1)*(4choose1)*(4choose1)*(4choose1)*(4choose1) ) /2598960 = ~3.95x10^(-4)
d)
(4choose3)*(4choose2) /2598960 = 1/108290

 

[Tedjn]

Let's look at (1) first. The number of ways to choose 2 of the 10 die to be rolled a 2 is 45, but that's not the total number of configurations in which exactly 2 die show a 2. Why not?

 

[sneaky666]

I don't get how to complete it. I am too lost...

 

[sneaky666]

anyone?

 

[sneaky666]

Let's look at (1) first. The number of ways to choose 2 of the 10 die to be rolled a 2 is 45, but that's not the total number of configurations in which exactly 2 die show a 2. Why not?

I don't under stand what i am doing wrong
i did
10x9 / 2x1 = 45
10x9 is because after i get one result there's 9 dice, and divinding it by 2! means i am removing all repeated sequences.