What's wrong with my reasoning? =( (1 Viewer)

[Shadowdude]

First off, I must say that I hate permutations and combinations. It sucks.

Now, if someone can tell me where I've gone wrong with my reasoning here, it'd be appreciated... (also there are 15 rep points in it for you):

Question: How many eight-card hands chosen from a standard pack have at least one suit missing?

"Solution": The number of eight-card hands chosen from a standard pack with at least one suit missing is the number of eight card hands chosen from a standard pack with one suit missing, plus those with two suits missing, plus those with three suits missing. It is impossible to have a hand with four suits missing.

Case 1: One suit missing

1. Choose the missing suit (e.g. hearts) ... 4 ways
2. Choose a card from one of the other suits (e.g. clubs) ... 13 ways
3. Choose a card from another different suit (e.g. spades) ... 13 ways
4. Choose a card from the last suit (e.g. diamonds) ... 13 ways
5. Choose 5 other cards from the remaining 36 with order not important and repetition not allowed ... C(36,5) ways

Sub-total: 4*13*13*13*C(36,5)

Case 2: Two suits missing

1. Choose two missing suits (e.g. hearts, diamonds) ... C(4,2) ways
2. Choose a card from one of the other suits (e.g. clubs) ... 13 ways
3. Choose a card from the remaining suit (e.g. spades) ... 13 ways
4. Choose 6 cards from the remaining 24 with order not important and repetition not allowed ... C(24,6) ways

Sub-total: C(4,2)*13*13*C(24,6)

Case 3: Three suits missing

1. Choose the suit that appears (e.g. hearts) ... 4 ways
2. Pick 8 cards from that suit with order not important and repetition not allowed ... C(13,8)

Sub-total: 4C(13,8)

"Answer": 4*13*13*13*C(36,5) + C(4,2)*13*13*C(24,6) + 4C(13,8) = 3 449 491 188

And that'd be a perfectly good answer, if it wasn't 3 212 764 698 over the correct answer.

Help?

 

[ohexploitable]

easier to subtract n(no suits missing) from n(no restrictions)

hahahaha disregard that i suck cocks

 

[bleakarcher]

^lol

 

[ohexploitable]

perms/combs and probability are my worst topics too

okay ummmm

one suit missing:
two suits missing:
three suits missing:

then multiply by 4 since there are 4 suits

total ways = 252350956

edit: which seems a bit off the correct answer

 

[Shadowdude]

Correct answer according to the back of the book is: "236 726 490"

or: 4C(39,8) - 6C(26,8) + 4C(13,8).

 

[deterministic]

The flaw in your first attempt comes from steps 4/5- where you choose other cards from the remainder. It means you will double count some cases. For example:
Suppose in case where hearts and diamonds are missing:
1) choose 10C and 10S
2) Choose 6 other cards to be 9S,9C,1S,2S,3S,4S
This is one combination.

Consider this combination
1) choose 9C and 9S
2) Choose 6 other cards to be 10S,10C,1S,2S,3S,4S
Although this is identical to the above, your method treats this as a different case. The correct answer below is based on the inclusion exclusion principle.
( 1 missing)-(2 missing)+(3 missing)
In this way, there is no double counting.

 

[Shadowdude]

http://answers.yahoo.com/question/index?qid=20080509222932AATwNCS

Wow, someone went to Yahoo Answers about it... weird.

The flaw in your first attempt comes from steps 4/5- where you choose other cards from the remainder. It means you will double count some cases. For example:
Suppose in case where hearts and diamonds are missing:
1) choose 10C and 10S
2) Choose 6 other cards to be 9S,9C,1S,2S,3S,4S
This is one combination.

Consider this combination
1) choose 9C and 9S
2) Choose 6 other cards to be 10S,10C,1S,2S,3S,4S
Although this is identical to the above, your method treats this as a different case. The correct answer below is based on the inclusion exclusion principle.
( 1 missing)-(2 missing)+(3 missing)
In this way, there is no double counting.

If it's based on the inclusion/exclusion principle, what are the 'sets' I have to consider? And... can you explain further? (I kinda missed a few perm and combs lectures in Discrete...)

 

[deterministic]

Its similar to how you split the cases in the first place:
1) Find all the cases where 1 suit is missing (this includes where 2 and 3 suits are missing):
- Choose a suit: 4
- Choose 8 cards from remaining 39
In the above case, you will notice the cases with 2 and 3 suits missing will be double counted. So lets subtract the 2 suit case:
2) Find the number of cases where 2 suits are missing (this includes where 3 suits are missing):
- Choose 2 suits: 4C2
- Choose 8 cards from remaining 26
So for the 3 suit case, we doubled counted in case (1), but we have also double counted in case (2), so subtracting (2) from (1) would result in the 3 suit case not being counted. So we just need to add it on (4 suits to choose 3 from, 13 cards to choose 8 from).
So overall:
4C1*39C8-4C2*26C8+4C3*13C8

Best way to picture it is in the form of a Venn Diagram

The sets you consider usually comes naturally, where the areas you double count are pretty obvious and can easily be calculated. For example, in counting the cases where 1 suit is missing, it is obvious there will be cases where more than 1 suit is missing which will be counted twice. In your original approach, the part you double counted is not obvious and hence not easy to calculate.

 

[Shadowdude]

So I drew a Venn diagram and went "Ohhhhhhhhhhhhh, so that's why".

Now I'll pretend I didn't read the solution so I can't 'cheat' so I can figure it out myself. Mmkay.

Now to solve the problem...