Suppose that two cards are dealt from a standard 52-card poker deck. let a be the event that the sum of the two cards is 8 (assume that aces have a numerical value of 1). how many outcomes are in a?

Case 1:  You have an Ace and a seven. 
There are 4 ways to pick 1 Ace out of 4. There are 4 ways to pick 1 seven out of 4.   > 4*4 = 16
There are 16 ways to be dealt an Ace and a seven.

Case 2:  You have a two and a six.
In the same way there are 4*4=16 ways to be dealt a two and a six.

Case 3: You have a three and a five.
In the same way there are 4*4=16 ways to be dealt a three and a five.

Case 4: You have 2 fours.
There are 6 ways to get 2 fours out of 4.  (4 choose 2)

Total number of outcomes = 16+16+16+6 = 54

Idon’t understand sorry

Same length is comparing because it’s saying its the same and not different