# Need to clarify the “At-least Concept” in Combination.

I managed to solve this question but I had some inquiries regarding the solution.

If two cards are chosen at random from a standard deck of playing cards, how many different ways are there to draw the two cards if at least one card is a jack, queen or a king?

Here is how I solved it:

Jack/Queen/King = $4 \times 3 = 12$ cards

Other Cards = $52-12 = 40$ cards

Now we can only pick two cards so:

• It can be either from the Jack/King/Queen so $\binom{12}{1}=12$

• Both cards can be from Jack/King/Queen so $\binom{12}{2} = 66$

• One card can be from the remaining stack (Non - jack,king or queen) so $\binom{40}{1}=40$

Now the only problem I have with this question is when getting the final value initially I was doing

$1$ from Remaining Cards $\times$ [ ($1$ from Jack/King/Queen) + 2(from Jack/King/Queen) ]= $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} )$

But The actual answer comes if we do the following

[$1$ from Remaining Cards $\times$ $1$ from Jack/King/Queen ] + 2(from Jack/King/Queen) ]= $( \binom{40}{1} \times \binom{12}{1}) + \binom{12}{2}$

I would really appreciate it if someone could clarify why do we do it the second way and not the first way ? Which part is added/multiplied to which part ? Is there an easier way to know how its done. Am I missing some important concept here ?

Edit : While trying to understand this I also looked up the definition of disjoint events which means "Two events are disjoint if they can't both happen at the same time" so then again here is what I did (Special here means Jack/king/Queen)

(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special)

$(\binom{40}{1} \times \binom{12}{1}) + (\binom{40}{1} \times \binom{12}{2}) )$

Which simplifies to $\binom{40}{1} \times ( \binom{12}{1} + \binom{12}{2} )$

So is the representation of the problem using

(1 from the 40 and 1 from special) or (1 from the 40 and 2 from special) wrong ?

-
If someone tells you he have at least one card is a jack, queen or a king then this means that it is not possible that he doesn't have a king or a queen or a jack (example: if he does not have a jack or a queen he must be holding a king) – Belgi Aug 6 '12 at 21:32
There are $\binom{52}{2}$ ways to choose two cards. There are $\binom{40}{2}$ ways to choose them so none is J, Q, or K. So an alternate form of the answer is $\binom{52}{2}-\binom{40}{2}$. In this case, probably a little harder than the way you used, but in other situations the idea of counting complement can save a lot of work. – André Nicolas Aug 6 '12 at 21:50
This seems rather obvious: the product term $\tbinom{40}1\times\tbinom{12}2$ represents "$1$ card from the remaining stack and both cards from Jack/Queen/King", or as you say it in the edit "1 from the 40 and 2 from special". But since only two cards are drawn this is not among the possibilities of the problem, so the term should not be present. A product represents two independent choices, and mutually exclusive (yet individually possible) events are never independent. – Marc van Leeuwen Aug 7 '12 at 9:27

If you expand your first answer, you have $$\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2} \cdot \binom{40}{1}.$$ Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.

A rule of thumb is to turn "at least" problems into several instances of "exactly".

For the problem at hand, "at least one jack/queen/king" translates into "either exactly one jack/queen/king or exactly two jack/queen/king".

The number of ways to get exactly one jack/queen/king and one other card is $$\binom{12}{1} \cdot \binom{40}{1}.$$ We multiply here because you choose a card from the jack/queen/king pile and a card from the "other" pile.

The number of ways to get exactly two from jack/queen/king is $$\binom{12}{2}.$$ Since these cases are disjoint (this is important), we simply add the results from the two cases. Thus, the number of ways to get at least one Jack, Queen, or King is $$\binom{12}{1} \cdot \binom{40}{1} + \binom{12}{2}.$$

-
thanks for the answer , could you explain what you meant by Notice in the term on the right, you are choosing a total of three cards, which does not count what you want.. – MistyD Aug 6 '12 at 21:38
@MistyD The term $\binom{12}{2} \cdot \binom{40}{1}$ instructs you to select two cards from the jack/queen/king pile and one card from the "other" pile, giving you a three-card hand (not a two-card hand, like the problem is asking for). – Austin Mohr Aug 6 '12 at 21:39
Though I partially get what you meant. Lets consider the correct way then $( \binom{40}{1} \times \binom{12}{1}) + \binom{12}{2}$ . Which I believe means 1 from remaining cards (which is 40) and 1 from J/K/Q , now what does + (mean here) . My mind keeps on asking me why we didn't multiply here instead of adding? – MistyD Aug 6 '12 at 22:05
Addition here corresponds to the union operation on sets; if the sets $A$ and $B$ have no elements in common (this is a critical restriction!) then $|A\cup B| = |A| + |B|$. In effect what you're saying is that 'all the combinations with at least one JQK' consists of 'all the combinations with exactly one JQK' and 'all the combinations with more than one JQK' (in this case there can be only two). – Steven Stadnicki Aug 6 '12 at 22:24
@MistyD You have two sheets of paper in front of you. The first lists all $\binom{40}{1} \cdot \binom{12}{1}$ hands having exactly one J/Q/K. The second lists all $\binom{12}{2}$ hands having exactly two J/Q/K. Together, the two sheets list all hands having at least one J/Q/K. Thus, to get the total count, you would want to add the number of hands in the first list to the number of hands in the second list, not multiply. – Austin Mohr Aug 7 '12 at 1:07

The easiest way to solve this problem is to note that subtracting the number of ways of selecting any two cards neither of which is a Jack, Queen, or King, from the number of ways of selecting any two cards, leaves with the number of ways to pick two cards of which at least one of them is a Jack, Queen, or King.

So,

$$\mbox{(Ways to select any 2 cards from the deck)} - \mbox{(Ways to select 2 cards where neither one is J/Q/K)} = \mbox{Ways to select 2 cards where at least one is J/Q/K}$$

$$\binom{52}{2} - \binom{40}{2} = 1326 - 780 = 546$$

-
Thanks for you answer but I am more interested in knowing why I am wrong.. – MistyD Aug 6 '12 at 22:29
@ladaghini This answer saves effort, and is more generally useful. Having a hard time making it work for picking at least 2 of a subgroup when picking 5 form the whole. – csga5000 Jul 14 '15 at 19:44
For my situation (15 balls, 8 red, 7 black, 5 chosen) the answer was: 15C5 - (7C4 * 8C1) - 7C5 Where C is combination operation, and the left/right numbers are n/r respectively. Total combinations - (combainations with 4 black 1 red) - (combaintions 5 black) – csga5000 Jul 14 '15 at 19:54