Probability is the study of how likely something is to happen. Permutations and combinations are two ways of counting how many different arrangements or selections can be made from a set of objects. In this article, we will review the basics of permutations and combinations, and see how they can help us solve problems involving probability.
A permutation is an arrangement of a set of objects where order matters. For example, if we have three letters A, B, and C, we can arrange them in six different ways: ABC, ACB, BAC, BCA, CAB, and CBA. Each of these arrangements is a permutation of the three letters.
To find the number of permutations of n objects, we can use the factorial notation. The factorial of a positive integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 5! = 5 � 4 � 3 � 2 � 1 = 120. We can also define 0! = 1 by convention.
The number of permutations of n objects is equal to n!. This is because we have n choices for the first object, n - 1 choices for the second object, n - 2 choices for the third object, and so on, until we have one choice for the last object. Multiplying these choices gives us n!.
For example, the number of permutations of 4 objects is 4! = 4 � 3 � 2 � 1 = 24.
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A combination is a selection from a set of objects where order does not matter. For example, if we have three letters A, B, and C, we can select two of them in three different ways: AB, BC, and CA. Each of these selections is a combination of the three letters.
To find the number of combinations of n objects taken r at a time, we can use the formula:
C(n,r) = n! / (r! (n - r)!)
This formula can be derived by dividing the number of permutations of n objects taken r at a time by the number of permutations of r objects. This is because each combination corresponds to r! permutations that have the same objects in different orders.
For example, the number of combinations of 4 objects taken 2 at a time is:
C(4,2) = 4! / (2! (4 - 2)!) = (4 � 3 � 2 � 1) / (2 � 1 � 2 � 1) = 6
Permutations and combinations can be used to count the number of possible outcomes in a probability experiment. For example, if we want to find the probability of getting a full house (three cards of one rank and two cards of another rank) in a five-card poker hand, we can use combinations to count the number of ways to get a full house and divide it by the total number of possible poker hands.
The total number of possible poker hands is C(52,5) = 52! / (5! (52 - 5)!) = 2598960.
The number of ways to get a full house is C(13,1) � C(4,3) � C(12,1) � C(4,2). This is because we have 13 choices for the rank of the three cards, 4 choices for their suits out of which we need to choose 3, 12 choices for the rank of the two cards out of which we need to choose one different from the first rank , and finally four choices for their suits out of which we need to choose two.
So, C(13,1) � C(4,3) � C(12,1) � C(4,2) = (13 � (4! / (3! (4 - 3)!)) � (12 � (4! / (2! (4 - 2)!))) = (13 � 4 � 12 � 6) = 3744.
Therefore, the probability of 8cf37b1e13