Cna Mac Excel do this calculation?

[angel]

I teach Bridge. A bridge hand has 13 cards, each card can be a value of
1 thru 13. Is there any calculation that I can do that will tell me the
total number of possible combinations? Thanks

 

[JE McGimpsey]

I teach Bridge. A bridge hand has 13 cards, each card can be a value of
1 thru 13. Is there any calculation that I can do that will tell me the
total number of possible combinations? Thanks

Note: since you're referring to bridge, I'll assume that suits matter.

One way to calculate the number of combinations is to recognize that the
first card in the hand is one of 52 possible cards. The second card is
one of 52-1, or 51 possible cards, therefore the possible permutations
(unique sequences) of 2 cards is

=52*51

or 2652 possible permutations. Note, however, that bridge hands don't
depend on the order the cards are dealt, so that the sequence [ace of
hearts, deuce of clubs] is equivalent to the sequence [deuce of clubs,
ace of hearts]. So the number of *combinations* of two cards is

=52*51/2

or 1326. Extending this, there will be

=52*(51/2)*(50/3)*(49/4)*...*(40/13)

possible combinations for 13 cards. If you're familiar with factorials
(e.g., 5 factorial, or 5! = 5 * 4 * 3 * 2 * 1), you can use this
shorthand:

# combinations = n! / (k! * (n - k)!)

where n is the total number of possible values and k is the number of
values chosen. So the number of possible bridge hands is

= 52! / (13! * (52 - 13)!)

In XL, Factorials are returned by the FACT function:

=FACT(52) / (FACT(13) * FACT(52-13)) ===> 6.35014E+11

But XL also has a function that does this for you - COMBIN():

=COMBIN(52, 13) ===> 635013559600

or a bit over 635 billion combinations.

 

[angel]

I cannot begin to tell you how much I appreciate your reply, especially
your very detailed explanation. My students are in for a very big
surprise! I completely forgot in my request that there are only 4
aces, 4 kings, etc., so my basic request was flawed. I hate to tell you
how long ago I was in High School when I would have known how to do
this.

I teach Bridge. A bridge hand has 13 cards, each card can be a value of
1 thru 13. Is there any calculation that I can do that will tell me the
total number of possible combinations? Thanks

Note: since you're referring to bridge, I'll assume that suits matter.

One way to calculate the number of combinations is to recognize that the
first card in the hand is one of 52 possible cards. The second card is
one of 52-1, or 51 possible cards, therefore the possible permutations
(unique sequences) of 2 cards is

=52*51

or 2652 possible permutations. Note, however, that bridge hands don't
depend on the order the cards are dealt, so that the sequence [ace of
hearts, deuce of clubs] is equivalent to the sequence [deuce of clubs,
ace of hearts]. So the number of *combinations* of two cards is

=52*51/2

or 1326. Extending this, there will be

=52*(51/2)*(50/3)*(49/4)*...*(40/13)

possible combinations for 13 cards. If you're familiar with factorials
(e.g., 5 factorial, or 5! = 5 * 4 * 3 * 2 * 1), you can use this
shorthand:

# combinations = n! / (k! * (n - k)!)

where n is the total number of possible values and k is the number of
values chosen. So the number of possible bridge hands is

= 52! / (13! * (52 - 13)!)

In XL, Factorials are returned by the FACT function:

=FACT(52) / (FACT(13) * FACT(52-13)) ===> 6.35014E+11

But XL also has a function that does this for you - COMBIN():

=COMBIN(52, 13) ===> 635013559600

or a bit over 635 billion combinations.