In a card game using a normal deck of playing cards containing 52 cards, 13 cards are dealt to a player.
a) how many different combinations of cards are possible in that players hands?
b) how many different hands are possible that consist ofnine spades and four clubs. Then leave t in factorial form.
for a i am thinking p(52,13)
and b i am not oo sure how to answer that.
For part a) of the question, you are correct in using the concept of permutations to determine the number of different combinations of cards in a player's hand.
The formula for permutations is P(n, r) = n! / (n - r)!, where n is the total number of items and r is the number of items chosen.
In this case, the player is being dealt 13 cards from a deck of 52 cards. Therefore, the number of different combinations is given by P(52, 13):
P(52, 13) = 52! / (52 - 13)!
= 52! / 39!
Now, let's calculate this value. Keep in mind that 52! means multiplying all the numbers from 52 down to 1, and 39! means multiplying all the numbers from 39 down to 1.
52! = 52 x 51 x 50 x 49 x 48 x ... x 1
39! = 39 x 38 x 37 x ... x 1
Canceling out the common terms between the numerator and denominator, we get:
P(52, 13) = (52 x 51 x 50 x 49 x 48 x ... x 40) / (13 x 12 x 11 x ... x 1)
After evaluating this expression mathematically, you will get the total number of different combinations of cards in a player's hand.
For part b) of the question, you need to determine the number of different hands that consist of nine spades and four clubs and express it in factorial form.
To approach this, we need to consider the number of ways we can choose 9 spades from the 13 available and 4 clubs from the 13 available. Using the concept of combinations, we can calculate this as:
C(13, 9) x C(13, 4)
Where C(n, r) = n! / (r! * (n - r)!)
C(13, 9) = 13! / (9! * (13 - 9)! )
= 13! / (9! * 4! )
Similarly,
C(13, 4) = 13! / (4! * (13 - 4)! )
= 13! / (4! * 9! )
Multiplying the two combinations together:
C(13, 9) x C(13, 4) = (13! / (9! * 4!)) x (13! / (4! * 9!))
Now, notice that the numerator and denominator of the two combinations are identical. Therefore, we can simplify this expression as:
C(13, 9) x C(13, 4) = (13! / 9!) x (13! / 4!)
Now, simplifying further:
C(13, 9) x C(13, 4) = (13 x 12 x 11 x 10) x (13 x 12 x 11 x 10 x 9 x 8 x 7 x 6)
Since this expression cannot be simplified any further, we can express the result as a factorial:
C(13, 9) x C(13, 4) = 13! / ((13 - 9)!) x 4!
Therefore, the number of different hands is 13! / 4!