Download Factorials: Definition, Illustrations, and Calculation and more Study notes Reasoning in PDF only on Docsity!
§9.FACTORIAL
Recall that (^) nPk, the falling factorial, stands for n(n−1)(n−2)... .(n−k+1). A special case occurs when k=n; in this case (^) nPn = n(n−1)(n−2)... .3.2. We will not write it as (^) nPn since have a special name and symbol for this product.
DEFINITION. The factorial of n is the number n(n−1)(n−2)... .3.2.1, to
be denoted henceforth by n! According to the results of our previous secIon, n! is the number of n‐lists of n different objects. In other words, n! = number of permutaDons of n disDnct objects.
Hence, three people can stand in line in 3! ways. Since 3! = 3.2.1 = 6, we have 6 different arrangements. We have 1! = 1 2! = 2 3! = 6 4! = 24 5! = 120, etc We complete our list by defining 0! = 1
ILLUSTRATIONS.
a) How many lists of length 5 from a pool of 0‐0‐1‐2‐3 if the two zeros are placed next to each other? SoluDon. To avoid misunderstandings, we can think of a bag with five balls, two numbered with 0, one with 1, one with 2 and one with 3.
Case 1: C _ _ _. The second spot can be occupied by B or G (but not by D) So two choices to fill the second spot. The third spot, likewise, can be filled in two ways. Finally the remaining spot is occupied in only one way. By the FCP our case 1 can be completed in 2 × 2 × 1 ways, that is, 4 ways. Case 2: _ C _ _ In this case D must occupy the rightmost posiIon (read the statement, please). So we must consider _ C _ D. This is completed by placing B and G, then in 2! ways. So case 2 takes 2 ways. Case 3: _ _ C _ Totally similar to case 2. So 2 more ways. Case 4: _ _ _ C Totally similar to case 1. So 4 more ways. We have considered all the possible arrangements, so we must add the numbers to get the total. That is, the answer is 12.
c) A bag contains five balls, numbered 0, 1 , 1, 2 and 3. Find in how many ways they can be arranged in a line if the first two balls must add up to 2. SoluDon. We list the possible arrangements: 2 0 _ _ _ :we must fill them with objects from the pool 1‐1‐3 (Careful here: they are not disInct elements) We have 3 ways, according to where the ball numbered 3 is placed) 0 2 _ _ _ as before. We have 3 more ways (none of them has been counted above, since the lists will differ in the first entry to begin with) 1 1 _ _ _ we count the permutaIons of 0, 2, 3 that is, 3! = 6 ways Having analyzed all the cases we conclude that there are 12 different arrangements saIsfying the condiIon of the problem.
Since the computaIon of factorials involve products, it is useful to introduce a symbol to simplify the notaIon. We denote by the product a 1 × a 2 ×... × an. Observe that k (the dummy variable) can be replaced by any other le]er (other than n, which indicates the upper limit for the range of k). Thus n! =
ILLUSTRATIONS.
a) 1 × 3 × 5 × 7 ×.... × 17 × 19 × 21
b) (−17) (−12) (−7) (−2) (3) (8) = 68, c) × ×..... × × = d) Compute
