Difficult Combination Question

[Samuelt1]

Hey guys if you could answer this by tonight that would be awesome. The question is: How many 8 card hands are possible if you must have a 3 of a kind and two pairs in the hand. The three of a kind and pairs may not be face cards the last card may be a face card. The pairs may not be the same i.e. no 4 of a kind.

What I have: (10c1)(4c3)(9c1)(4c3)(8c1)(4c3)(40c1)

I am not sure if this is right, i don't know if it should be just (9c2) or (9c1)(8c1) for the pairs.

Thanks !

 

[stapel]

Hey guys if you could answer this by tonight that would be awesome. The question is: How many 8 card hands are possible if you must have a 3 of a kind and two pairs in the hand. The three of a kind and pairs may not be face cards the last card may be a face card. The pairs may not be the same i.e. no 4 of a kind.

To clarify, just as the two pairs may not be parts of four-of-a-kind, would it be correct to assume that no pair could also be part of a three-of-a-kind? That is, the sets must be distinct? Thank you!

 

[samuel123t]

To clarify, just as the two pairs may not be parts of four-of-a-kind, would it be correct to assume that no pair could also be part of a three-of-a-kind? That is, the sets must be distinct? Thank you!

My account somehow got deleted… but yes thats correct they must be distinct

 

[Ishuda]

Now admittedly I don't always do too well on some types of permutations/combinations but it seems to me that you can do it the following way: Assume face cards are the Jack, Queen, and King.
You have 40 for the first number [40 non face cards]
You have 3 for the second number [matches the first number]
You have 2 for the third number [matches the first two numbers]
You have 36 for the fourth number [no face card and doesn't match the first three numbers]
...
Then take into account duplicates because order doesn't matter [we need permutations].

 

[Ishuda]

Shouldn't that be 52; if 1st = face, then other 7 can't = face ; no?

Don't think so: "...The three of a kind and pairs may not be face cards the last card may be a face card. ..."

 

[pka]

The question is: How many 8 card hands are possible if you must have a 3 of a kind and two pairs in the hand. The three of a kind and pairs may not be face cards the last card may be a face card. The pairs may not be the same i.e. no 4 of a kind.
What I have: (10c1)(4c3)(9c1)(4c3)(8c1)(4c3)(40c1)

Some of that is incorrect.

$\displaystyle \dbinom{10}{1}\dbinom{4}{3}\dbinom{9}{2}\left[\dbinom{4}{2}\right]^2\dbinom{40}{1}$

 

[Ishuda]

Disagree...OP means (I think) "the remaining card";
8 cards drawn, then arranged in ascending order (example):

12 1 4 4 7 4 7 1 : 1 1 4 4 4 7 7 12

Problem says "The three of a kind and pairs may not be face cards"
TRIPS
-That leaves 40 cards, i.e. there are 12 face cards in the deck, for the first draw. The second and third card match the denomination of the first card for the trips.
PAIR ONE
-Since you can not use the remaining card of that denomination for the pair draw, you have 36 cards left from the 40 because "The three of a kind and pairs may not be face cards" and you have already used 4 of them (three in the draw and one 'dead card'. To match that 4th card, you have 3 choices of the denomination left.
PAIR TWO
-That kills that denomination for the first pair, i.e. "The pairs may not be the same i.e. no 4 of a kind." So you are left with 32 cards [the pair can't be face cards leaves 40, killed 4 for the trips leaves 36, killed 4 for the first pair leaves 32] and then, to match (to make a pair) you have three choices.
LAST CARD
-Now we come to the last card [start with 52 but killed 4 with the trips leaves 48, killed 4 with the first pair leaves 44, killed 4 with the second pair, leaves 40].

Put it together to get
40 * 3 * 2 * 36 * 3 * 32 * 3 * 40, i.e. 1 1 1 2 2 3 3 J

 

[pka]

[1] Agree...that was never in dispute...

[2] that's for a total of 99,532,800; Pka's is 2,073,600

Btw, I'm not picking sides or calling anything correct or incorrect;
I'm trying to understand/follow what is going on...

The question is: How many 8 card hands are possible if you must have a 3 of a kind and two pairs in the hand. The three of a kind and pairs may not be face cards the last card may be a face card. The pairs may not be the same i.e. no 4 of a kind.

What I have: (10c1)(4c3)(9c1)(4c3)(8c1)(4c3)(40c1)
I am not sure if this is right, i don't know if it should be just (9c2) or (9c1)(8c1) for the pairs.

@Dennis. Please read the OP.
(10c1)(4c3)(9c1)(4c3)(8c1)(4c3)(40c1) is the same as
$\displaystyle \dbinom{10}{1}\dbinom{4}{3}\dbinom{9}{1}\dbinom{8}{1}\left[\dbinom{4}{2}\right]^2\dbinom{40}{1}=149299200$[/QUOTE]

Please note that answer is twice the answer that I gave, which is, of course, correct? WHY?

 

[Ishuda]

[1] Agree...that was never in dispute...

[2] that's for a total of 99,532,800; Pka's is 2,073,600
Soooo....what can I say?:cool:

Btw, I'm not picking sides or calling anything correct or incorrect;
I'm trying to understand/follow what is going on...

Denis,
Why do you think I started out with "Now admittedly I don't always do too well on some types of permutations/combinations ...". I'm willing to be convinced. Besides, that 99 532 800 is before duplicates are removed and that's generally where I get caught up.

Take a simple example of choosing any two cards from a deck of 52. First card is 52 choices and second card is 51 choices so there are 2652 ways to chose, right? Well, maybe; is the A of Spades, King of Hearts considered the same as the King of Hearts, Ace of Spaces? If the answer to that question is no, then the 2652 is correct and we are talking about permutations. However, if the answer is yes, they are considered the same, the number has to be reduced by the number of 'duplicates' and we are talking about combinations. In the simple example here, the ways you can arrange the two cards is 2!, so divide the 2652 by 2! to get 1326 ways.

In that sense, my 99 532 800 would have to be divided by 8!. However, that does not give a whole number so that must be incorrect. Where is it wrong? I don't know. That's why I didn't go any further and just said we had to remove the duplicates. Maybe I only have to divide by 4! since we only have 4 'sets'; 1 trips, 2 pairs, and 1 single? Or maybe the 4 sets and the arrangement of the two pairs, i.e. 4! * 2!. That latter would actually make my answer and pka's answer the same as the difference between the permutation number of 99 532 800 and pka's 2,073,600 is 48 = 4! * 2!

Edit to correct the spelling of "pka's" - sorry 'bout that pka

 

[pka]

Take a simple example of choosing any two cards from a deck of 52. First card is 52 choices and second card is 51 choices so there are 2652 ways to chose, Where is it wrong? I don't know. That's why I didn't go any further and just said we had to remove the duplicates.

The mistake is an easy one to correct. Often permutation and combinations are confused. A permutation is about order; a combination is about content. When questions about possible hands of cards are involved, that is a question about content not order. If a hand has one three-of-a-kind and two pairs, we understand there is no overlapping. In this question there are ten ways to select which denomination the triple will be and four ways to select the content.
There are thirty six, $\displaystyle \binom{9}{2}$, ways to select the denominations for the two pairs and $\displaystyle \binom{4}{2}=6$ ways to select the content of each pair. In the case of the OP, he/she and you chose the first pair and its content and then chose the second pair and its content. ORDER was introduced at that point; the count was doubled.

 

[Ishuda]

The mistake is an easy one to correct. Often permutation and combinations are confused. A permutation is about order; a combination is about content. When questions about possible hands of cards are involved, that is a question about content not order. If a hand has one three-of-a-kind and two pairs, we understand there is no overlapping. In this question there are ten ways to select which denomination the triple will be and four ways to select the content.
There are thirty six, $\displaystyle \binom{9}{2}$, ways to select the denominations for the two pairs and $\displaystyle \binom{4}{2}=6$ ways to select the content of each pair. In the case of the OP, he/she and you chose the first pair and its content and then chose the second pair and its content. ORDER was introduced at that point; the count was doubled.

It isn't a matter of a mistake. It is a matter of how one counts. I tend to prefer to do the count by permutations (order does matter) and then, if combinations are wanted, reduce the permutations by duplicates (order doesn't matter). Maybe I should change how I do it and it would become easier as it is that reduction that sometimes gets me in 'more complicated counting'.