Practice problems for these concepts can be found at:
 Probability Solved Problems for Beginning Statistics
 Probability Supplementary Problems for Beginning Statistics
Many of the experiments in statistics involve the selection of a subset of items from a larger group of items. The experiment of selecting two letters from the four letters a, b, c, and d is such an experiment. The following pairs are possible: (a, b), (a, c), (a, d), (b, c), (b, d), and (c, d). We say that when selecting two items from four distinct items that there are six possible combinations. The number of combinations possible when selecting n from N items is represented by the symbol and is given by
 (4.15)
_{N}C_{n}, C(N, n), and are three other notations that are used for the number of combinations in addition to the symbol .
The symbol n!, read as "n factorial," is equal to n × (n – 1) × (n  2) × . . . × 1. For example, 3! = 3 × 2 × 1 = 6, and 4! = 4 × 3 × 2 × 1 = 24. The values for n! become very large even for small values of n. The value of 10! is 3,628,800, for example.
In the context of selecting two letters from four, N! = 4! = 4 × 3 × 2 × 1 = 24, n! = 2! = 2 × 1 = 2 and (N – n)! = 2! = 2. The number of combinations possible when selecting two items from four is given by = , the same number obtained when we listed all possibilities above. When the number of items is larger than four or five, it is difficult to enumerate all of the possibilities.
EXAMPLE 4.27 The number of five card poker hands that can be dealt from a deck of 52 cards is given by = 2,598,960. Notice that by expressing 52! as 52 × 51 × 50 × 49 × 48 × 47!, we are able to divide 47! Out because it is a common factor in both the numerator and the denominator.
If the order of selection of items is important, then we are interested in the number of permutations possible when selecting n items from N items. The number of permutations possible when selecting n objects from N objects is represented by the symbol , and given by
 (4.16)
_{N}P_{n}, P(N, n), and (N)_{n} are other symbols used to represent the number of permutations.
EXAMPLE 4.28 The number of permutations possible when selecting two letters from the four letters a, b, c, and d is = 12. In this case, the 12 permutations are easy to list. They are ab, ba, ac, ca, ad, da, bc, cb, bd, db, cd, and dc. There are always more permutations than combinations when selecting n items from N, because each different ordering is a different permutation but not a different combination.
EXAMPLE 4.29 A president, vice president, and treasurer are to be selected from a group of 10 individuals. How many different choices are possible? In this case, the order of listing of the three individuals for the three offices is important because a slate of Jim, Joe, and Jane for president, vice president, and treasurer is different from Joe, Jim, and Jane for president, vice president, and treasurer, for example. The number of permutations is = 10 × 9 × 8 = 720. That is, there are 720 different sets of size three that could serve as president, vice president, and treasurer.
When using EXCEL to find permutations and combinations go to any cell and enter the EXCEL function =COMBIN(number, number—chosen) to find number of combinations. Enter the EXCEL function =PERMUT(number, number—chosen) to find the number of permutations. The answer to Example 4.27 using EXCEL is =COMBIN(52,5), which returns 2,598,960, and the answer to Example 4.29 is =PERMUT(10,3), which returns 720.

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