It's me again. I just wanted to ask how you find the number of possible subsets. For example, how do I find out how many 5 card subsets (order irrelevant) are in a 52-card deck or how many 2 card subsets (again, order irrelevant) are there in say, a 20 card deck? Any help with this problem would be appreciated.
"I just wanted to ask how you find the number of possible subsets. For example, how do I find out how many 5 card subsets (order irrelevant) are in a 52-card deck ?"
The definition goes something like this: The number of all possible combinations of r objects (for example 5 cards) taken out of a total of n objects (for example 52 cards) is denoted in math as nCr and is given by the formula
which means in your example
52C5 = (52!) / [(52-5)!*5!] = (52!) / (47!*5!)
Where 5! equals 5*4*3*2*1 and 52! equals 52*51*...*2*1. Simplifying with 47!, the above gives
= 52*51*50*49*48 / 5*4*3*2*1 = 2,598,960
which of course is the number of all possible hands in 5-card draw and, thus, an extremely important number in the minds of all poker players who don't have a life.
OK, I tried that but when I tried to find the number of 2 card subsets in a deck of 52 cards I got something like 1335. That can't be right, so am I missing something?
52!/50!*2! = 52*51/2 = 1326.
You asked for subsets of two for 20 cards. Work it out to see that the correct answer is 190.
Don
"I tried that but when I tried to find the number of 2 card subsets in a deck of 52 cards I got something like 1335. That can't be right, so am I missing something?"
The correct number for that one is 1,326, as Don Schlesinger pointed out, but I guess you feel "that can't be right" because it may seem like too high a number. If that is so, would you feel such a high number would be OK for 50-card subsets? I guess it would, given the "enormity" of a 50-cared subset. But please note the number of all possible combinations of 2 items out of a set of 20 items :
Items in subset # of Combinations
1 20
2 190
3 1,140
4 4,845
5 15,504
6 38,760
7 77,520
8 125,970
9 167,960
10 184,756
11 167,960
12 125,970
13 77,520
14 38,760
15 15,504
16 4,845
17 1,140
18 190
19 20
20 1
Actually, I thought it too low for a 52 card subset, since the number of 5 card subsets in a 52 card deck is in the millions I was thinking there should be more 2 card subsets than 5. Like in a 5 card subset, there is 1 5 card combination but 10 2 card subsets. Is there a midway point when the number of 5 and 2 card subsets is equal, and then subsequently there are more 5 card than 2 card subsets? Please pardon my ignorance.
"Is there a midway point when the number of 5 and 2 card subsets is equal, and then subsequently there are more 5 card than 2 card subsets?
I don't understand.
When you have a specific number of items to choose from, say for example 52, then the number of items you wanna put in each "subset" gives you a precise number for all the possible combinations.
Picking every time 5 items out of 52, gives you 2,598,960 possible combinations. (And I did not look up that one !)
Picking every time 2 items out of every 52, gives you 1,326 possible combinations. (As Don pointed out.)
What is the "midway point" you have in mind ?
What are you doing hanging out at a website about a game that you're not legally allowed to play?
Just curious.
A time honored tradition that I myself participated in.
6 years from now, when you are starting college, and got a good idea for a business, you might find some investment type backing from some of the people here.
Keep working on the web page.
Well, I read a lot and became interested in casino gambling about 1.5 years ago, and I am fascinated by the mathematics behind it. And I have virtually nothing else to do.
What other interests do you have?
I was 86'ed by my local Boy Scout Troop for running after-taps poker games on camping trips.
I used the money I made to buy my first guitar.
Disclaimer: I am not condoning or recommending that minors set up illegal gambling operations.
Just mainly mathematics, American History, true crime, HTML, and good music, like classical or Ki' Holau.
I started the same way. Read the Preface to BJA, p. xvii.
Don
OK, I am having another problem with it. My friend has a problem that is:
There are 4 parking spaces and 3 cars. How many different ways can all three cars go into the 4 spaces? He said it was 36, the answer key said 6. I counted and got 12. I used the Subset equation, and got 6. What's up with that? Did I screw something up along the line? Please let me know. Thanks.
"There are 4 parking spaces and 3 cars. How many different ways can all three cars go into the 4 spaces? He said it was 36, the answer key said 6. I counted and got 12. I used the Subset equation, and got 6. What's up with that?"
The 4 parking spaces are A, B, C and D, while the 3 cars are 1, 2 and 3. In every try, you are obliged to leave 1 parking space empty - but consider that you are parking there a phantom car, which I call car 4. You can see that cars 1-2-3 parked at A-B-C (leaving D empty) is the same as cars 1-2-3-4 parked at A-B-C-D. And 1-2-3 parked at A-B-D (leaving C empty), is the same as 1-2-4-3 parked at A-B-C-D.
The number of different set-ups is 4*3*2*1 = 24. Here they are :
A B C D
1 2 3 4
1 2 4 3
1 3 2 4
1 3 4 2
1 4 2 3
1 4 3 2
2 1 3 4
2 1 4 3
2 3 1 4
2 3 4 1
2 4 1 3
2 4 3 1
3 1 2 4
3 1 4 2
3 2 1 4
3 2 4 1
3 4 1 2
3 4 2 1
4 1 2 3
4 1 3 2
4 2 1 3
4 2 3 1
4 3 1 2
4 3 2 1
Does order of the cars matter?
if order matters, then let 1,2,3 equal unique cars and 0 represent an empty space.
0123 1023 1203 1230
0132 1032 1302 1320
0312 3012 3102 3120
0321 3021 3201 3210
0213 2013 2103 2130
0231 2031 2301 2310
24 possible unique arrangements.
if it is just a question of car, no car
0111
1011
1101
1110
4 possible arrangements.
Since order does not matter with the combination equation,
(n!) / [(n-r)!*r!]
4 parking spaces = n taken 3 cars at a time = r is
(4!) / [1! * 3!] = (1 * 2 * 3 * 4) / 1 * 1 * 2 * 3 = 4 / 1
or 4 combinations.
Is there a methematical equation that is order sensitive?
OK, thanks, now that you went through it I realize I forgot that there were 3 cars, yet I used 2 in the equation... Silly me. Now, I was wondering, is there a mathematical equation that is order sensitive?
how many ways can n objects be ordered?
n!
In this case you had four parking spaces, three with a unique car, one empty (which is considered unique also). So 4 parking spaces can be arranged 4! ways, or 4 * 3 * 2 * 1 = 24 different arrangements.
