In how many ways can each of the following be drawn from a standard deck of 52 cards

1. Brainstorm: documenting my attempts at solving the question. Justifying why I cam up with my solution even if it was wrong.
2. Solution: detailed walkthrough of the actual solution to the problem. Sometimes accompanied by diagrams.
3. Final Solution: Skip to this section to see the final answer.
4. Conclusion: What I learned.

Brainstorm

Seems like a simple question on the surface, but it took me a few days to wrap my head around this concept. A full house consists of 3 cards of one value and the remaining two of another value (i.e. JJJQQ). I initially approached the problem from the perspective that we must initially choose from the 52 total cards to find the number of total permutations, and then divide by the number of arrangements among the two subgroups — the group of 3 and the group of 2. This would correspond to the following formula:

This would result in

This clearly doesn’t work since for starters, we are not arranging 52 cards, and secondly, the n_k need to add to n and 3+2 does not equal 52. My next attempt was

Which is what I settled on before checking the textbook solution. Why did I settle on this? Well, the example prior to this one involved _ _ _ _ _ so for the first spot, it would be 52 possibilities. 52 _ _ _ _. Would this then leave 51 possibilities for the second spot? No. Once we select the first card, then we have simultaneously selected the card value, of which there are 3 remaining in the deck (4 total). The tough part about this question was the distinction between selecting from 52 cards in the beginning, to selecting from only 13. At first I couldn’t wrap my head around why we were selecting from 13 objects rather than the total 52. All prior questions involved choosing/arranging from a set number of distinct objects. But in this case, the objects we are selecting from are not distinct. That is, if we begin the problem by selecting from 52 overall cards, we treat each card as a distinct object, and therefore rule out the possibility of a full house in the first place! But wait a minute, can’t we use the formula n!/n1!*n2!*…*nk! ? This was my next idea. However this too doesn’t work since it treats all n objects as distinct. This formula works in the case where we are organizing n distinct objects into non-distinct groups. That is, say we have 5 paintings and want to sort 3 into one group and 2 into another, we can use this formula to do so. But in the case of a deck of cards we can’t use this formula since not all 8 cards are distinct. Moreover, we aren’t arranging 8 cards, we are only arranging 5 cards.

Solution

The problem can be visualized as 5 items needing to be arranged among 5 slots, where there are two distinct groups of objects, and the objects within each group are non-distinct. So _ _ _ _ _ could be JJJQQ Where the Jack’s are one distinct group, the Queen’s are the other distinct group, and the Jack’s are indistinguishable from one another (non-distinct) just as the Queen’s are also non-distinct. It seems cumbersome to use the distinct/non-distinct terminology but I find it helps when determining what formulae to use.

We can boil the problem down further. There are

ways to organize the 3 Jack’s from 4 total Jack’s, and

ways to organize the two Queen’s. We multiply these results to determine that there are 24 possible ways to organize 3 Jack’s and 2 Queen’s, which is one of many unique Full House arrangements. There are a few questions which could arise at this step:

1. Why do we use 4C3 and 4C2 instead of 4P3 and 4P2?

The order within the two respective groups does not matter. We aren’t counting the number of arrangements of the 3 Jack’s for example, as these cards are indistinguishable from one another in the context of this question. The arrangement: Jack of hearts, Jack of spades, Jack of clubs is equivalent to Jack of spades, Jack of hearts, Jack of clubs, and so on, since each unique arrangement among the Jack’s does not constitute a unique Full House arrangement. What about JQJQJ or JQQJJ or JQJJQ etc? Again these are all still the same Full House arrangement therefore we don’t need to count the unique arrangements among the 5 cards. This gives us the distinction between permutation and combination. The permutation 4P3 yields the same result for reasons I will discuss in another post, but 4P2 = 12 whereas 4C2 = 6. Therefore for the Queen’s we are double-counting the arrangements where the Queen’s positions’ vary, which again does not constitute a unique Full House arrangement, just a permutation of that unique arrangement.

2. Why do we multiply the results?

Everything in Combinatorics (to my current knowledge/experience) boils down to a tree diagram that represents the number of ways to perform an operation. Therefore, when we’re thinking about whether to add or multiply results it makes sense to consider how each operation would influence the corresponding tree diagram that describes all possible outcomes.

Logically each step is a multiple of the prior step. If we select a Queen of heart’s (top branch) then there are 3 ways to select the next card, then 2 ways and so on. This is a multiplicative expansion. The number of branches at each step is a constant factor. 4 options for the first branch, 3 options for each branch thereafter, and then 2, and then 1. It becomes clear that to count the final number of branches, we can simply multiple 4*3=12. The same logic holds when counting the total number of ways to choose the JJJQQ Full House. There are 4 ways to choose 3 Jack’s from 4 (so 4 options for the first branch) then for each of those 4 ways, there are 6 ways to choose 2 Queen’s from 4. It becomes the same counting problem. We simply multiply 4*6=24 since there are the same number of branches at each step.

Moving on, we know that for the Full House arrangement JJJQQ there are 24 ways to choose this arrangement. Note that this is different from there being 24 ways to arrange JJJQQ since we are not counting arrangements, we are counting the number of ways to choose this specific arrangement. Given that this is just one of many Full Houses, we should extend this logic to all other Full Houses. Essentially, the Tree Diagram of the solution will have 24 branches for each node, so we have started backwards.

Now the problem becomes: how many ways can we select two card values from 52 cards? This tripped me up initially, but after some thought, it makes sense that we are selecting from 13 values and not 52. If we select from 52 values, we treat all 52 values as distinct from one another, which in the confines of this question is not accurate. This is where these problems can get confusing; we need to change the definition of distinctness to suit the problem we are trying to solve. Sure, the 52 cards are distinct if we look at them from outside the question, but for the purposes of selecting Full Houses, we need only concern ourselves with the card values to differentiate among them. Therefore there are 13 ways to select the value of the card for the first 3 slots. Then, for each of these 13 cards, there are either 4C3 or 4C2 ways to arrange among them. Notice the two distinctions I make here: “for each” means that this problem can be thought of as a tree diagram, as for each node there are X ways to perform an action, and therefore our answer is multiplicative. “There are either” is also an important distinction, as this indicates that order matters. When order matters, we care about the ways the objects are arranged and not just how the items are selected. Specifically, JJJQQ is distinct from QQJJJ when we are counting the number of ways to select the two values.

Therefore, there are 13 ways to select the first value, and 12 ways to select the second value. Again we note that for each of the 13 ways to select the first value, there are either 4C3 or 4C2 ways to choose the cards from that value. Similarly, for each of the 12 ways there are to select the second value, there are either 4C3 or 4C2 ways to choose the cards from that value. Again, I highlight the distinctions because the wording is crucial to understanding this problem, and creating a framework that can extend to future problems. This wording does not distinguish whether the first suit is counting 4C3 or 4C2. Therefore it counts both possibilities: JJJQQ and QQQJJ since if Jack is selected as the first suit, we must count both possibilities where there are 3 Jack’s and 2 Jack’s and vice versa for Queen’s. This gives us our final solution.

Final Solution

Notice how we can frame this solution in several ways since the values are multiplicative so their order in the solution does not matter, but it could change the way we interpret the result. The above solution shows that there are 13 ways to select the first value, 12 ways to select the second value, and 4C3*4C2 ways to arrange each unique Full House that arises.

This solution shows that there are 13 ways to select the first card, and for each of those options, there are 4C3 ways to choose the 3 cards chosen from 4. Then there are 12 ways to select the next value, and 4C2 ways to choose the 2 cards from 4.

We can continue changing the order to come up with different rationales for our solution but all are equally valid.

Conclusion

This concludes my first ever Math/Statistics blog post! I’m not sure how helpful this will be for anyone else to read as my description is pretty convoluted, but this is a first step for me in terms of simplifying my thought process so that I can become better at teaching this type of material.