why is this true?

[trickslapper]

Four cards are dealt at random without replacement form an ordinary deck of 52 cards.

If it is known that the four cards have different face values, what is the proability that the first card is a spade?

I have the answer and the solution but i can't make sense of it, the solution says that A is the event that the four cards have different face values and that B is the event that the first card is a spade. Ok no problem there but here is where i'm confused:

According to the solution guide

Event A can occur 13[sub:m00srggy]4[/sub:m00srggy] * 4[sup:m00srggy]4[/sup:m00srggy] ways------> How did they get this ????

and Event A Intersect B can occur: 13* 12[sub:m00srggy]3[/sub:m00srggy] 4^3 ways------->and how did they get this????

I understand the idea and what the problem wants but i don't understand where these numbers came from, can someone explain this to me?

 

[galactus]

Four cards are dealt at random without replacement form an ordinary deck of 52 cards.

If it is known that the four cards have different face values, what is the proability that the first card is a spade?

I have the answer and the solution but i can't make sense of it, the solution says that A is the event that the four cards have different face values and that B is the event that the first card is a spade. Ok no problem there but here is where i'm confused:

According to the solution guide

Event A can occur 13[sub:u6muissq]4[/sub:u6muissq] * 4[sup:u6muissq]4[/sup:u6muissq] ways------> How did they get this ????

First, we have to choose 4 from the 13 different values: A,2,3,4,5,6,7,8,9,10,J,Q,K.

Thus, \(\displaystyle \binom{13}{4}\)

But, of those 4 cards, each can have one of 4 suits. 4 choices for the first, 4 choices for the second, and so on.

Thus, 4^4 ways.

\(\displaystyle \binom{13}{4}\cdot 4^{4}=183,040\)

and Event A Intersect B can occur: 13* 12[sub:u6muissq]3[/sub:u6muissq] 4^3 ways------->and how did they get this????

Choose 1 of the 13 spades, then the 3 remaining cards can be chosen by choosing 3 from the other 12 values, then there are 4 choices for suits of those 3 non-spade cards. 4 for the first, 4 for the second, and 4 for the third. Thus, \(\displaystyle 13\cdot \binom{12}{3}\cdot 4^{3}=183040\)



I understand the idea and what the problem wants but i don't understand where these numbers came from, can someone explain this to me?[/quote]

 

[trickslapper]

damn, i guess i should have realized that 13[sub:2nc1id9f]4[/sub:2nc1id9f] meant "13 choose 4" thanks though!