Choose 6 cards from a normal deck

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karmayogi
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Topic: Choose 6 cards from a normal deck
PostTue Sep 22, 2009 5:48 am
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20. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

My answer is ((13^4)48C2). I want to understand, what's wrong with my logic?
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PostTue Sep 22, 2009 10:09 am
Whats the Original answer to this question?
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PostTue Sep 22, 2009 10:09 am
Karma,

After a little confusion, I found the source of this question and concluded that there is a typo for (B). Techincally, you are correct.

Remember 48C2 = 48!/2!46! = 24*47, so that would lead to (B) as it should have been written:

b. (13^4) x 24 x 47

Note that (A) is wrong because 48*47 assumes order matters. In this case, order is irrelevant, so you have to use 48C2 instead of 48*47.
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PostTue Sep 22, 2009 12:21 pm
karmayogi wrote:
20. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

My answer is ((13^4)48C2). I want to understand, what's wrong with my logic?
Very tough question (probably well beyond the scope of the GMAT). I wanted to confirm my thoughts before posting, so I asked the teachers on our province-wide listserve for Math teachers. The responses (including one from the math department at the University of British Columbia) confirm my suspicians about the solution 13^4 x 48C2.

Here's my reasoning:

First, 13^4 x 48C2 is too large.
I understand the reasoning here: We take each of our 4 suits and select one card from each suit (13 ways for each suit).
Now that we have fulfilled our obligation to have one of each suit, we then select two more cards from the remaining 48 (48C2)
However, this allows us to count some selections more than once.
In the first step, I can select an ace of spades (to fulfill my "all suits" obligation) and then later select a two of spades (when selecting 2 additional cards).
Conversely, I could select a two of spades (to fulfill my "all suits" obligation) and then later select an ace of spades (when selecting 2 additional cards). So, the above approach would count these separately, when they are the same selection.

To avoid this, I took a somewhat long approach.
I recognized that the question's restrictions yields two cases:
1) One card from each of two suits, and 2 cards from each of two other suits.
# of options = 4C2 (select the two suits that will have two cards from them) TIMES 13C2 (select 2 cards from one of the 2-card suits) TIMES 13C2 (select 2 cards from the other 2-card suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from the other 1-card suit). We get 4C2 x 13C2 x 13C2 x 13 x 13

2) One card from each of three suits, and 3 cards from one suit.
# of options = 4C1 (select the one suit that will have three card from it) TIMES 13C3 (select 3 cards from that suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from the other 1-card suit). We get 4C1 x 13C3 x 13 x 13 x 13

So, adding both cases, we get (4C2 x 13C2 x 13C2 x 13 x 13) + (4C1 x 13C3 x 13 x 13 x 13).

If we "simplify" this, we get 13^4(6^3 + 88)
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PostThu Sep 24, 2009 4:58 pm
Brent Hanneson wrote:
karmayogi wrote:
20. In how many ways can one choose 6 cards from a normal deck of cards so as to have all suits present?
a. (13^4) x 48 x 47
b. (13^4) x 27 x 47
c. 48C6
d. 13^4
e. (13^4) x 48C6

My answer is ((13^4)48C2). I want to understand, what's wrong with my logic?
Very tough question (probably well beyond the scope of the GMAT). I wanted to confirm my thoughts before posting, so I asked the teachers on our province-wide listserve for Math teachers. The responses (including one from the math department at the University of British Columbia) confirm my suspicians about the solution 13^4 x 48C2.

Here's my reasoning:

First, 13^4 x 48C2 is too large.
I understand the reasoning here: We take each of our 4 suits and select one card from each suit (13 ways for each suit).
Now that we have fulfilled our obligation to have one of each suit, we then select two more cards from the remaining 48 (48C2)
However, this allows us to count some selections more than once.
In the first step, I can select an ace of spades (to fulfill my "all suits" obligation) and then later select a two of spades (when selecting 2 additional cards).
Conversely, I could select a two of spades (to fulfill my "all suits" obligation) and then later select an ace of spades (when selecting 2 additional cards). So, the above approach would count these separately, when they are the same selection.

To avoid this, I took a somewhat long approach.
I recognized that the question's restrictions yields two cases:
1) One card from each of two suits, and 2 cards from each of two other suits.
# of options = 4C2 (select the two suits that will have two cards from them) TIMES 13C2 (select 2 cards from one of the 2-card suits) TIMES 13C2 (select 2 cards from the other 2-card suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from the other 1-card suit). We get 4C2 x 13C2 x 13C2 x 13 x 13

2) One card from each of three suits, and 3 cards from one suit.
# of options = 4C1 (select the one suit that will have three card from it) TIMES 13C3 (select 3 cards from that suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from a 1-card suit) TIMES 13 (select 1 card from the other 1-card suit). We get 4C1 x 13C3 x 13 x 13 x 13

So, adding both cases, we get (4C2 x 13C2 x 13C2 x 13 x 13) + (4C1 x 13C3 x 13 x 13 x 13).

If we "simplify" this, we get 13^4(6^3 + 88)
Why is the following reasoning wrong:

You have six spots to fill.

1st spot: 52 choices (doesnt matter what card/suit you choose)

2nd spot: 52-1-12 = 39 choices (subtract the card you chose from the first spot plus all the cards of the same suit)

3rd spot: 39-1-12 = 26 choices (same reasoning)

4th spot: 26-1-12 = 13 choices (any card from the last remaining suit that has not been used yet)

5th spot: 52-4 = 48 choices (do not matter which of the remaining cards you choose)

6th spot: 48-1= 47 choices (any of the last remaining cards)

Giving you:

52C1 x 39C1 x 26C1 x 13C1 X 48c1 X 47c1 =

52*39*26*13*48*47 =

1,546,406,784 different ways to pick the cards out of the deck so that you have a hand of 6 cards with at least 1 of each suit.
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PostFri Sep 25, 2009 7:49 am
benjiboo wrote:
Why is the following reasoning wrong:
You have six spots to fill.

1st spot: 52 choices (doesnt matter what card/suit you choose)

2nd spot: 52-1-12 = 39 choices (subtract the card you chose from the first spot plus all the cards of the same suit)

3rd spot: 39-1-12 = 26 choices (same reasoning)

4th spot: 26-1-12 = 13 choices (any card from the last remaining suit that has not been used yet)

5th spot: 52-4 = 48 choices (do not matter which of the remaining cards you choose)

6th spot: 48-1= 47 choices (any of the last remaining cards)

Giving you:

52C1 x 39C1 x 26C1 x 13C1 X 48c1 X 47c1 =

52*39*26*13*48*47 =

1,546,406,784 different ways to pick the cards out of the deck so that you have a hand of 6 cards with at least 1 of each suit.
I'll begin with the first part (where you select the first 4 card such that each card is a different suit). The error here is similar to the error employed when selecting the last 2 cards.

In your approach, several identical selections are counted more than once (24 times actually).
In your approach, I can select a jack of hearts for my first card. Then, I remove the remaining hearts from the deck and select the next card. Let's say I select an ace of spades. Then we remove the remaining spades from the deck and select my next card, say a five of diamonds. Finally, I remove the remaining diamonds and select my last card, say a ten of clubs.

So, my selection is jack of hearts, ace of spades, five of diamonds and ten of clubs.

In another selection, I select the ace of spades first (then remove the remaining spades), then the ten of clubs (then remove the remaining clubs), then the five of diamonds (remove diamonds) and then the jack of hearts.

In your approach, you would count this selection as one that is different from your first selection, even though it is the same.
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PostWed Nov 18, 2009 1:53 am
This is soooooooooo interesting question. Smile

With a little bit of simplification, I was able to prove that there is no right answer provided in the choices.

Then, let me share the approach I used. Here, the problem is the number of card deck is way too large for human being to understand complicated calculation. So, let's simplify this!

Lets assume that there is a card deck with only 6 cards! and we choose 4 cards from them.

The composition of a new card deck
1,2 heart
3,4 spade
5,6 diamond

Assume that except the composition and number of card, the other conditions for the question is same as the given question.

Here, if answer (b) is correct approach, there must be 2*2*2*3C1 = 24 ways to meet the condition.

However, if you acutally list all the possible outcomes, you will get the list like this

(1,3,5) 2 4 6 ---------3 ways
(1,3,6) 2 4 -----------2
(1,4,5) 2 6 -----------2
(1,4,6) 2 --------------1

(2,3,5) 4 6 -----------2
(2,3,6) 4 -------------1
(2,4,5) 6 -------------1
(2,4,6) --------------- 0

If you add them up, you will get 12...which is only half of 24!

So we can cleary see that there is double counting problem. I will update this if I come up with more detailed information about calculating the double counting matter.
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