ryan selects one card from a deck of 52 different card. the first card is not replaced before joe selects the second one. in how many different ways can they select the two card?
To find the number of different ways Ryan and Joe can select two cards from a deck of 52 cards, we can use the concept of combinations.
When selecting two cards from a deck without replacement, the order in which the cards are chosen does not matter. Therefore, we need to find the number of combinations of choosing 2 cards out of 52.
The formula to calculate combinations is:
C(n, r) = n! / (r!(n-r)!)
Where:
n is the total number of items (52 cards in this case)
r is the number of items to be chosen (2 cards in this case)
! denotes the factorial operation (e.g., 5! = 5 x 4 x 3 x 2 x 1)
Plugging the values into the formula:
C(52, 2) = 52! / (2!(52-2)!)
= 52! / (2!50!)
Calculating the factorials:
52! = 52 x 51 x 50 x ... x 2 x 1
2! = 2 x 1
50! = 50 x 49 x 48 x ... x 2 x 1
Simplifying the equation:
C(52, 2) = (52 x 51 x 50 x ... x 2 x 1) / [2 x 1 x (50 x 49 x 48 x ... x 2 x 1)]
Canceling out common terms:
C(52, 2) = (52 x 51) / (2 x 1)
Calculating the expression:
C(52, 2) = (52 x 51) / 2
= 2652
Therefore, there are 2652 different ways in which Ryan and Joe can select two cards from the deck.