How To Tell When Order Matters Or Not I have encountered a problem involving combinatorics: 

My solution to it was $(4\cdot3\cdot2)+(5\cdot3\cdot4)+(6\cdot5\cdot4)$. 
The textbooks solution to it, however, was 

I would understand the solution if order didn't matter, but I don't think, from the problem hints, that order doesn't matter. 
Can someone please explain this to me? What about the problem shows that order doesn't matter? Thank you. 
 A: Deciding whether the order matters or not in this case is more of an English problem than a math problem. It's whether the phrases "select cards at random" and "the number of selections" refer to things where order matters or not.
In this case it turns out that order didn't matter, but I see no way of being certain of that from the problem statement itself. Nothing in there mentions whether Grace cares which card is first, second and third, or if she only cares about which cards she ends up with.
That being said, if you were to calculate the probability of ending up with such a hand, it doesn't matter which interpretation you go with. You'll get the same answer either way. The same cannot be said if repetitions are allowed. If repetitions are allowed, and order matters, then a hand of $1111$ is as common as a hand of $1234$, while if order doesn't matter, then a hand of $1234$ is 24 times more likely than a hand of $1111$.
A: Usually in a selection with some constraints the order doesn't matter. But context is decisive.
This especially goes for a hand of cards, where in virtually any card game only the contents of your hand matters, not in what order you drew those cards.
That being said, it is slightly ambiguous and the problem should probably state it explicitly.
A: The problem only tells you the the largest number in four cards is either 5, 6 or 7. It doesn't tell you in which position it is.
Shuffling the same selection of four cards doesn't affect the larger value among them. So order doesn't count.
A: a) When order matters, the total number of ways to select four cards is $9*8*7*6$:
the 4-tuples are distinguished by content and/or by order.
b) If the order does not matter then it will be $\binom{9}{4}$: the 4-tuples are distinguished only by content.
Our universe is given by all the sets $\{1 \le x < y < z < w \le 9 \}$.    
In case a) the number of ways to select, e.g. $(x,y,z,5) \; |\, max(x,y,z)=4$ is $4*3*2$ which is what you computed.
But the $5$ can be in whichever position, so actually it is $4*(4*3*2)$.
The probability to select four cards with a max of $5$ then is $4*(4*3*2)/(9*8*7*6)$.
In case b) the number of ways to select $(x,y,z,5)\; |\, x<y<z \le 4$ is $(4*3*2)/3! = \binom{4}{3}$.
The probability is 
$$
\binom{4}{3} \;\mathop /\limits_{} \; \binom{9}{4}
 = {{4 \cdot 3 \cdot 2} \over {3!}}{{4!} \over {9 \cdot 8 \cdot 7 \cdot 6}} = 4{{4 \cdot 3 \cdot 2} \over {9 \cdot 8 \cdot 7 \cdot 6}}
$$
which is the same as in case a).
So the conclusion is that it is always important to specify the "universe" of the events (conditions) being considered.
In your case that was not clearly specified.
