Problem Solving

Randolph has a deck of 12 playing cards made up of only 2 suits of 6 cards each. Each of the 6 cards within a suit has a different value from 1 to 6; thus, there are 2 cards in the deck that have the same value.

Randolph likes to play a game in which he shuffles the deck, turns over 4 cards, and looks for a pair of cards that have the same value. How many such combinations are possible?

A. 240
B. 960
C. 120
D. 40
E. 5760


I tried the above problem using Combinations, and I got the correct answer which is A.

Here's the method : 2* (6C1 * 1C1 * 5C2) + (6C2 * 2C1 * 4C1) = 240 {First part - Choose 1 card from the first suit, remaining two cards from the second suit; Second part - Choose 2 from each} {1C1 and 2c1 represent the same numbered card}


However, I crashed while using Permutations. Permutations - 12*1*10*8 = 960. (crash ) I don't know how I can arrive at 240 using this method. Really confused.

I tried a couple of other examples:

LEt's say there are only 4 cards of two suits each.

Combinations: 2* (4C1*1C1*3C1) + (4C2*2C1*2C1) = 48 {Same logic - choose 1 from the first suit; remainign three from the second suit; Second part - Choose 2 from each. }
Using permutations - 8* 1*6*4= 48*4...(crash )


I see that in both the examples, the "Combinations" number is "Permutations/4"...Is there a rule to quickly relate Permutations with Combinations?

I know that nCr = nPr/r!; However, I believe that this theorem is not applicable here.

Correct?

Your calculation of 12*1*10*8 = 960 permutations only includes the scenario in which the first two cards form a pair, and the other two don't.

There are additional scenarios. If P denotes a pair card and N denots a non-pair card, there are 6 different scenarios (anagrams of PPNN):

PPNN
NPPN
NNPP
PNNP
PNPN
NPNP

Each scenario has 960 permuations in it for a total of 6*960 permutations that have two cards forming a pair and two cards not forming a pair.

Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination.

The number of combinations is then 6*960/24 = 960/4 = 240.

The most efficient approach to this problem is neither combinations nor permutations but using the slot principle (multiply possibilities for each slot):

total combinations = (ways to choose two pair cards) x (ways to choose two non-pair cards)

ways to choose two pair cards = 6 because there are 6 pairs total when the order of the cards in the pair does not matter

ways to choose two non-paired cards = 10*8/2!=40 because 10 cards are left after a pair is chosen and then 8 cards are left that won't match the first chosen non-pair cards. The factorial 2! corrects for the fact that the card order does not matter in the non-pair.

So total number of combinations is 6*40 = 240.

tutorphd wrote:Your calculation of 12*1*10*8 = 960 permutations only includes the scenario in which the first two cards form a pair, and the other two don't.

There are additional scenarios. If P denotes a pair card and N denots a non-pair card, there are 6 different scenarios (anagrams of PPNN):

PPNN
NPPN
NNPP
PNNP
PNPN
NPNP

Each scenario has 960 permuations in it for a total of 6*960 permutations that have two cards forming a pair and two cards not forming a pair.

Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination.

The number of combinations is then 6*960/24 = 960/4 = 240.

Awesome. Can you please explain why you did "Since all the cards are different (either by value or by suit), the theorem you mentioned does apply and 4!=24 permutations correspond to a single combination."

The theorem applies only when you are selecting from a set of different objects. If some of the objects are not different/distinguishable, not all combinations will correspond to the same number of permutations, and the number of permutations and combinations are not related so easily by a simple division.

Take for example 3 throws of a fair coin. The possible outcomes are 8 permutations, like HTH etc.

The combination {H, T, H} corresponds to 3 permutations: THH, HTH, HHT.
The combination {H, H, H} corresponds to just 1 permutation: HHH.
So you can't find all combinations just by dividing all permutations by 3!.

Tutorphd,
Thanks for your analysis. However, I didn't follow this part "Since all the cards are different (either by value or by suit) and 4!=24 permutations correspond to a single combination. " in your post.

There are four cards, out of which two of them are same (=pair). Hence, a combination could look like AABC => there are 4!/2! permutations and not 4!. Why did you divide by 24?

Thoughts please?

Because the permutation calculation of 960 favorable permutations in each scenario treats each card as distinguishable by number or suit or both - that is implicit in the calculation used to get 960. You don't correct there for 2 in suit A and 2 of suit B "being the same".

That is why, when you convert the number of permutations to number of combinations, you have to keep treating all the cards as different. For example, if the two suits are denoted by A and B, the combination {2A, 2B, 1A, 3A} corresponds to 24 permutations of the 4 different elements.

you can also do the following:
with 12 cards you can form 6 pairs of identical cards: P1, P2, P3, P4, P5, P6
Once you have one pair of identical cards, you are left with 10 cards made up of only 2 suits of 5 cards each.
with these 10 cards you can make 45 combinations of 2 cards 10C2=45.
The 45 combinations contain 5 pairs of identical cards and 40 pairs of different cards.
In order to have one pair in a set of 4 cards, you need to pick one pair from 6 and one pair from 40 (one pair from the identical cards, and one pair from the non identical cards)

Total number of combinations[spoiler]=6*40=240[/spoiler][/spoiler]

tutorphd wrote:Because the permutation calculation of 960 favorable permutations in each scenario treats each card as distinguishable by number or suit or both - that is implicit in the calculation used to get 960. You don't correct there for 2 in suit A and 2 of suit B "being the same".

That is why, when you convert the number of permutations to number of combinations, you have to keep treating all the cards as different. For example, if the two suits are denoted by A and B, the combination {2A, 2B, 1A, 3A} corresponds to 24 permutations of the 4 different elements.

I didn't folllow this. In calculating 960, we did "12*1*10*8" instead of "12*11*10*9" - why are you saying that the cards were assumed distinguishable? Only three cards were assumed distinguishable. I feel that I am missing something very fundamental....

Thoughts / help?