# 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.

• Well, I guess if you have a pack of cards and you shuffle it, it's still the same pack of cards. Therefore the order doesn't matter. – Matti P. Oct 17 '18 at 10:15

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 $$1111$$.

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.

• Comically badly worded question - particularly amusing is the phrase 'each card displays one positive integer without repetition from this set' :) it's almost like the output of a bot fed elementary combinatorics questions! – Mehness Oct 17 '18 at 10:24

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.

• How do you know it doesn't matter? What if she is using these cards to play a game where it matters which card is drawn first? We may be interested to know what the highest card is regardless of order, but we could still care about the order. – Arthur Oct 17 '18 at 10:30
• @Arthur Well, I don't. Since the only request was about the larger of 4 numbers, I thought it was safe to assume it was the only thing that mattered. I see your point, though. An explicit statement would have definitely be better. – francescop21 Oct 17 '18 at 14:50

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 $$(x,y,z,5)\; |\, x is $$(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.