There are 52 cards in a standard deck. if you draw a card, do not replace it, and then draw 4 more cards, one at a time without replacement, how many possible outcomes are there?
I don't get it...Thank you
To determine the number of possible outcomes when drawing cards without replacement, you can use the concept of combinations.
First, let's break down the problem:
When you draw the first card, you have 52 options since there are 52 cards in the deck.
After drawing the first card, there are 51 cards remaining in the deck. So, when drawing the second card, you have 51 options.
Similarly, after drawing the second card, there are 50 cards remaining, so you have 50 options for the third card.
For the fourth card, you have 49 options, and for the fifth card, you have 48 options.
To calculate the total number of outcomes, you multiply all the options together:
52 * 51 * 50 * 49 * 48 = 311,875,200
Therefore, there are a total of 311,875,200 possible outcomes when drawing a card without replacement and then drawing four more cards.
Keep in mind that this calculation assumes every order of drawn cards is considered a unique outcome. If the order doesn't matter, you would use combinations instead of permutations.
your first card could be 52 different ones, leaving 51 different ones to draw for you second, then 50 for the third and 49 for the fourth, and finally 48 for the last.
so what is 52(51)(50)(49)(48)
= ...