Respuesta :
Answer:
a) 182 possible ways.
b) 5148 possible ways.
c) 1378 possible ways.
d) 2899 possible ways.
Step-by-step explanation:
The order in which the cards are chosen is not important, which means that we use the combinations formula to solve this question.
Combinations formula:
[tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
In this question, we have that:
There are 52 total cards, of which:
13 are spades.
13 are diamonds.
13 are hearts.
13 are clubs.
(a)Two-pairs: Two pairs plus another card of a different value, for example:
2 pairs of 2 from sets os 13.
1 other card, from a set of 26(whichever two cards were not chosen above). So
[tex]T = 2C_{13,2} + C_{26,1} = 2*\frac{13!}{2!11!} + \frac{26!}{1!25!} = 182[/tex]
So 182 possible ways.
(b)Flush: five cards of the same suit but different values, for example:
4 combinations of 5 from a set of 13(can be all spades, all diamonds, and hearts or all clubs). So
[tex]T = 4*C_{13,5} = 4*\frac{13!}{5!8!} = 5148[/tex]
So 5148 possible ways.
(c)Full house: A three of a kind and a pair, for example:
4 combinations of 3 from a set of 13(three of a kind ,c an be all possible kinds).
3 combinations of 2 from a set of 13(the pair, cant be the kind chosen for the trio, so 3 combinations). So
[tex]T = 4*C_{13,3} + 3*C_{13.2} = 4*\frac{13!}{3!10!} + 3*\frac{13!}{2!11!} = 1378[/tex]
So 1378 possible ways.
(d)Four of a kind: Four cards of the same value, for example:
4 combinations of 4 from a set of 13(four of a kind, can be all spades, all diamonds, and hearts or all clubs).
1 from the remaining 39(do not involve the kind chosen above). So
[tex]T = 4*C_{13,4} + C_{39,1} = 4*\frac{13!}{4!9!} + \frac{39!}{1!38!} = 2899[/tex]
So 2899 possible ways.
