We often find the probability of an event by counting the number of elements in a simple sample space.

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outig Methods We ofte fid the probability of a evet by coutig the umber of elemets i a simple sample space. Basic methods of coutig are: Permutatios ombiatios Permutatio A arragemet of objects i a defiite order is called a permutatio of the objects. Suppose the objects are all distict. How may differet permutatios are there?! permutatios. Uless stated otherwise, selectios are made at radom, ad all! permutatios are equally likely to occur. Each permutatio occurs with probability.! How may differet ways are there to fill the 9 battig positios from 9 players? Fid the probability that Smith will be just after Joes. 9 Homework We fill the first four battig positios from 9 players. Fid the probability that both Joes ad Smith are amog the four ad Smith is just after Joes.

2 ombiatios The umber of differet ways of selectig r objects out of distict objects is called the umber of combiatios of thigs take r at a time, deoted by r or. r For permutatios, the order matters: ( abc,, ) ( acb,, ). { abc,, } { acb,, } For combiatios, oly the set matters:. r! ( r)! r! r r 0 I radom selectio of distict objects, all combiatios are equally likely. A NBA team has 2 players. How may ways ca the coach choose the startig five? What is the probability that Smith will be oe of the startig five? Assume the coach picks players at radom. 5 PiclSmith [ ]. 2 A Ituitive Pick 5 amog 2. 5 PiclSmith [ ]. 2 7 PexclSmith [ ]. 2

3 Permutatios of No-Distict Objects { abb} How may differet ways are there of orderig three letters,,? Each letter is used oce ad oly oce. Write the two b's as b ad b2. There are 3! permutatios: a b a b b a b b b a b b 2 2 2 2 2 2 b b b a b a However every sequece is idetical to aother sequece if b b 2. There are 3! 3 differet way of orderig them. 2! a b b b a b b b a The problem is idetical to that of where to place a. 3 3 ways. The sample space is simple. I geeral, How may differet ways are there of orderig 2, K objects of item, objects of item 2, objects of item K? ( + + + ) 2!!! 2 The sample space is simple. K K!

4. ( aaaaabbbcc) { aaaaabbbcc} How may differet ways are there of orderig te letters,,,,,,,,,? Each letter is used oce ad oly oce. Fid the prob of,,,,,,,,,. There are 0! ways. 5!3!2! The sample space is simple, ad 5!3!2! Paaaaabbbcc [,,,,,,,,, ]. 0! A Ituitive Aswer : Pick letters oe by oe. 5 P ( a) 0 54 P ( a, a) 0 9 5 432 P ( aaaaa,,,, ) 0 9 8 7 6 5 432 32 P ( aaaaabbb,,,,,,, ) 09876 543 5 432 32 2 5!3!2! P ( aaaaabbbcc,,,,,,,,, ) 0 9 8 7 6 5 4 3 2 0! For ay permutatio, e.g., 532 42 32 5!3!2! P ( abcabcaaab,,,,,,,,, ) 0 9 8 7 6 5 4 3 2 0!

5. Partial Permutatio of No-Distict Objects How may differet ways are there of selectig ad orderig two letters out of { aaaaabbbcc} ( a a) the te letters,,,,,,,,,? Each letter is used at most oce. Fid P,. There are ie ways: aa ab ac ba bb bc ca cb cc However the sample space is ot simple: 5 4 aa ( ) 0 9 ab ac ba bb bc ca cb 2 cc ( ) 0 9 I geeral, No geeral rule. The probability of a outcome depeds o the experimet.

6. Partial Permutatio of No-Distict Objects There are 0 red balls ad 0 blue balls. We are goig to pick 0 balls ad the order them. How may differet ways are there of orderig the 0 balls? Is the sample space simple? osider 0 cells, each of which is to be filled with a ball: Sice we have 0 red balls ad 0 blue balls, each cell ca be filled with a red or blue ball. 0 There are 2 differet ways. [ ] 0 P 0 red balls 2 if the sample space were simple. However the sample space is ot simple because: P [ 0 red balls] 0 9 8 20 9 8 0!0! 20! 20 0 ituitive? 6 5.4 0 << 2 0 We expect the probability will be highest for the evet i which the te cells are filled with five red ad five blue balls. Homework There are 0 red balls ad 0 blue balls. We are goig to pick balls ad the order them. How may differet ways are there of orderig the balls? Is the sample space simple?

7 ombiatios of No-Distict Objects There are 25 pairs of socks i a box. 5 pairs are red ad 0 pairs are blue. We pick seve pairs. What is the probability that there are exactly three red pairs amog the seve? Thik as if the objects are distict. Name the 25 socks as R, R,, R, B, B,, B so that the 25 socks are all distict. The there are 25 7 For pickig 3 red pairs, Therefore P[ 3 red socks] 2 5 2 0 ways of selectig 7 socks, ad the sample space is simple. there are ways of selectig 3 red socks, ad there are ways of selectig 4 blue socks. 5 3 0 4 5 3 0 4 25 7. A Ituitive Select 7 socks oe by oe:. We wat to pick exactly three red socks. There are 7 3 combiatios of cells to which we ca place the three red socks. 5 4 3 0 9 8 7 Each outcome has the same probabiliy. 25 24 23 22 2 20 9 For example, R R R B B B B R R B R B B B 5 4 3 0 9 8 7, 25 24 23 22 2 20 9 5 4 0 3 9 8 7, 25 24 23 22 2 20 9 7 3 0 5 9 4 8 7 3 B R B R B B R. 25 24 23 22 2 20 9 5 4 3 0 9 8 7 7! 8! 5! 0! 5! 0! 8!7! 53 0 4 25 24 23 22 2 20 9 4!3! 25! 2! 6! 2!3! 6!4! 25! 25 7

8 outig Permutatios of No-Distict Objects usig ombiatios Permutatios of o-distict objects ca be solved through combiatios There are 3 red shoes ad 2 blue shoes. We order the 5 shoes. How may differet ways are there of orderig the 5 shoes? What is the probability that the first two shoes are blue? Usig permutatios of o-distict objects, we ca see there are 5! 0 3!2! differet ways of orderig the five shoes. A alterative method is as follows. The problem is equivalet to selectig two cells for placig blue shoes amog the five cells. B B. There are 5 2 0 differet ways, ad the sample space is simple. P[ first two shoes are blue ] P[ BBRRR]. 0 A Ituitive 2 blue blue P [ first two shoes are blue] 5 shoes to choose from 4 shoes to choose from 0 Homework There are 0 red balls, 0 blue balls, ad 0 white balls. How may differet ways are there of orderig the 30 balls? Is the sample space simple?