That's how I used to solve things like this and if you look closely, it's the same thing.
51! = 51 x 50! = 51 x 50 x 49! = ...
I'll explain using smaller numbers.
Let's say you have ABCDEF, 6 letters.
The total number of combinations is 6!.
Now, A and B should be adjacent.
(AB)CDEF
Make as if A and B were stuck together; they form one item, for a total combination of 5!
But now, A and B could be either in the order AB or BA, so, you multiply by 2, the number of combinations with A and B together; 2 x 5!
Probability becomes $\displaystyle \dfrac{2\times 5!}{6!}$
Sorry for my previous mistakes... it's 10 past midnight now, and I'm really tired. I'll answer tomorrow if there is still some unclear things.
@Sneaky,
I don't get the last line of your statement. If you tried to picture a single ace and put the other one to the left or to the right of it, I think you might have overlooked the 'edges' (top and bottom of the deck) where the ace has only one neighbouring site?
ok but i have one more question
Suppose we roll 3 fair die, what is the chances that 2 match and one is different.
i put for this
there are 216 outcomes
chance of getting a number on die1 = 6/6
chance of getting the same number as on die1 in die2 = 1/6
chance of getting a different number as die1 in die3 = 5/6
6/6 * 1/6 * 5/6 = 5/36
Is this right?
Let's try it this way:
(i) There are 50! possibilities to get the aces of spades in the first and the ace of clubs in the second slot. Right?
(ii) There are 50! possibilities to get the aces of clubs in the first and the ace of spades in the second slot. Right?
(iii) There are 51 possible positions of the pair in the deck. Ok?
(iv) There are 52! possible arrangements of cards in the deck in total.
(v) So in the end we have 2 * 51 * 50! possibilities to have the pair from 52! possibilities in total. This gives a probability of 2 * 51 * 50! / 52! = 2 / 52.
Which step is unclear to you?
A triple with exactly one pair looks like $\displaystyle (X,X,Y)$
Now the $\displaystyle X's$ can have six values and the $\displaystyle Y's$ can have five.
That is $\displaystyle 30$. But each triple can be arranged is $\displaystyle 3$ ways. That is $\displaystyle 90$ in all.