MATH 1823 Honors Calculus I Permutations, Selections, the Binomial Theorem

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1 MATH 823 Hoors Calculus I Permutatios, Selectios, the Biomial Theorem Permutatios. The umber of ways of arragig (permutig) obects is deoted by! ad is called factorial. I formig a particular arragemet (or permutatio) we have choices for the first obect. Give a particular choice for the first obect, we have ( ) choices for the secod, ad so o. Evetually we have ( )( 2)... (3)(2)() possible choices i formig a arragemet. Thus there are ( )( 2)... (3)(2)() possible arragemets of obects.! ( )( 2)... (3)(2)() Note that!. It is a accepted covetio to defie 0!. For example, there are 2! 2 arragemets of the letters a, b. Here they are: ab, ad ba. There are 3! 6 arragemets of the three letters a, b, c. They are: abc, acb, bac, bca, cab ad cba. Q]... Write out all the arragemets of the 4 letters a, b, c, ad d. How may are there? There are 4! 24 of them. Here they are: abcd, abdc, acbd, acdb, adbc, adcb, bacd, badc, bcad, bcda, bdac, bdca, cabd, cadb, cbad, cbda, cdab, cdba, dabc, dacb, dbac, dbca, dcab, dcba Q2]... Write dow the followig umbers (use a calculator to help you) 5!, 6!, 7!, 8!, 9! ad 0!. 5! 20, 6! 720, 7! 5040, 8! 40320, 9! , 0! As you have guessed from the last exercise above, the factorials grow extremely rapidly. There is a pretty result, due to a dead guy called Stirlig, which says that! grows like 2π ( e ) This remarkable growth has surprisig practical implicatios. Some of these implicatios are the reaso why some compaies ivest huge fiacial resources i developig fast algorithms which give reasoably accurate results rather tha focusig o algorithms which give the best possible results, but at huge costs i computig time. Take the travelig salesma problem for istace. The problem here is oe of optimizatio (actually miimizig travel costs). Suppose a compay has to sed a salesma to visit differet cities i the mid-wester ad wester U.S. Which order should the salesma visit the cities i order to miimize the total travel costs/time? The aswer seems to be very simple: you are told by a travel aget the costs of travelig betwee all possible pairs of cities. Now you simply list all possible routes ad compute their total roud-trip costs, ad the select the roud-trips with the lowest costs. This clearly gives the best possible results, but what about the computig time? You thik about this for a secod: there are choices for the first city, the ( ) choices for the secod, etc. Q3]... How may total roud trips are there? There are a total of! roud trips. So you decide to program a computer to list all the possible roud-trips, ad to do all the additios quickly to get totals for each roud trip. Suppose the computer ca compute 000 roud trip totals per secod.

2 Q4]... How log will it take it to deal with 20 cities? How log will it take it to deal with 25 cities? It will take 20!/(000)(60)(60)(24)(365.25) years (the term accouts for leap years) to deal with 20 cities. This is a total of 77,094,02.48 years. It will take 25!/(000)(60)(60)(24)(365.25) years to deal with 25 cities. This is a total of 49,52,058,594,920 years, or about the time it takes to determie the outcome a presidetial electio!

3 Selectios. The umber of ways of selectig (or choosig) a group of r obects (for example a committee) from a group of obects is deoted by ( r) which is read choose r. How do we fid a formula for ( r)? Well, first look at all the possible ordered lists of r obects that ca be selected from obects. There are choices for the first member of the list. Give that choice, there are ow ( ) choices for the secod member of the list, ad so o. Thus there is a total of ( )... ( r + ) possible ordered lists of r thigs that ca be selected from a group of thigs. Now, these lists ca be combied ito groups of size r!, where all the lists i a particular group are ust arragemets of a give selectio of r thigs from the origial group of. Thus, the total umber of choices of r thigs from the origial group of thigs is give by ( )... ( r + ) r! We ca tidy this up by multiplyig above ad below by ( r)... (2)() to get r! r!( r)! We take this as a defiitio of ( r) eve if r 0. We remember to use the covetio that 0!. Q5]... Do this aalysis explicitly to determie the umber of selectios of 3 thigs from the group of five letters a, b, c, d ad e. That is, write dow all the ordered lists of size three which ca be obtaied from these 5 letters (o repetitios withi a give list of course!). The group your lists together if they ivolve the same three letters. How big is each group? How may groups of lists do you have? There are 0 groups each cotaiig 6 lists. This is a total of 60 (5)(4)(3). Here they are abc abd abe acd ace ade bcd bce bde cde acb adb aeb adc aec aed bdc bec bed ced bac bad bae cad cae dae cbd cbe deb dec bca bda bea cda cea dea cdb ceb dbe dce cab dab eab dac eac ead dbc ebc ebd ecd cba dba eba dca eca eda dcb ecb edb edc ( Q6]... Compute the followig umbers: 0 ) ( 0, ) ( 0, ) (, 2 ) ( 0, 2 ) (, 2 ) ( 2, 3 ) ( 0, 3 ) (, 3 ) ( 2, 3 ) ( 3, 4 ) ( 0, 4 ) (, 4 ) 2, ( 4 ) ( 3, 4 ) 4. Put your aswers dow i rows, labeled by the upper umber 0,, 2, 3, ad 4. Do you recogize the result? What is it? The result is Pascal s triagle. Here it is. ) ) ( ) 2 ) ( 2 ) ( 2 2) 3 ) ( 3 ) ( 3 ) ( 3 2 3) 4 ) ( 4 ) ( 4 ) ( 4 ) ( )

4 Q7]... Prove that ( r r). Give a ituitive iterpretatio of this fact. We have r! ( r)!( ( r))!! ( r)!r! r ad we re doe! This result should be ituitively true, sice every time you make a selectio of r thigs from a group of thigs, you automatically make a selectio of r thigs (the remaiig or complemetary thigs). Distict selectios have distict complemets. Thus, the umber of ways of selectig r thigs is the same as the umber of ways of their r complemets. Q8]... Show that ( ( 0) ad ) are always equal to. Now, we will get the rows of Pascal s triagle, provided we ca prove that the ( ( r) add together to give other r) ust like i Pascal s triagle. Prove that + + r r + r + Fially, Q7] above ow cofirms our observatio of the symmetry i the rows of Pascal s triagle. We have 0! 0!( 0)!!! ad so ( ( ) 0) too. Now for the additio formula + r r +! r!( r)! +! (r + )!( (r + ))!! r!( r)! +! (r + )!( r )!!(r + ) r!(r + )( r)! +!( r) (r + )!( r )!( r)!(r + ) (r + )!( r)! +!( r) (r + )!( r)!!(r + ) +!( r) (r + )!( r)!!(r + + r) (r + )!( r)!!( + ) (r + )!( r)! ( + )! (r + )!(( + ) (r + ))! + r + This also has a ituitive iterpretatio. Here it is. Suppose you wat to select a committee of r + people from a roomful of + people. We kow that the total umber of possible committees is ( + r+). You might like to kow how may of those committees cotai a particular perso (let s call him Paddy!) i the room. Well we ca create a committee of r + people which cotais Paddy, by simply choosig Paddy, ad the choosig r other people from the remaiig people i the

5 room. There is a total of ( ( r) ways of doig this. Thus there are r) committees of r + people which cotai Paddy. What about the Paddy-free committees. Well you simply create these by tellig Paddy to leave the room, ad the choosig the full committee of r + from the remaiig people. There are obviously ( ) ( r+ ways of doig this. Thus there are r+) Paddy-free committees of r + people. Now a give committee either cotais Paddy or is Paddy-free. Thus, ( + r+) must be the sum of ( r ad r+). Doe!

6 Biomial Theorem. This theorem tells you that the ( r) are precisely the coefficiets of a r b r i the expasio of (a + b). Usig the summatio otatio developed i class (ask me if you missed this!) it says (a + b) a b 0 We wot give a borig proof as is the usual case at this stage, but istead will focus o a ituitive uderstadig. Our expressio cosists of a product of bracketed terms as show: (a + b)(a + b)(a + b) (a + b) Note that the term a appears by takig a a out of each bracketed term ad multiplyig them together. We get a ab by takig a a out of the first bracketed term ad b s out of all the remaiig bracketed terms. We get aother ab by takig the a from the secod bracketed term ad b s from the others: it appears as bab... b. Q9]... How may terms (product of legth cosistig of a s ad b s i some order) are there altogether? How may of these give rise to a ab? List these explicitly. There are 2 biary words (that s strigs to you computer sciece maors!) of legth. We see this by otig that to create such a word, we have two choices for the first letter, two for the secod, ad so o. Of these exactly ( ) give rise to the term ab. Here they are writte out explicitly: abbb... bb babb... bb bbab... bb. bbb... ba The ( ) comes from the fact that we have to choose oe slot from i which to place the a ad fill the remaiig slots with b s. Q0]... How may terms give rise to a a r b r? Remember, we have to choose r bracketed expressios from amog the list of bracketed expressios from which to take a s, ad the we take b s from the remaiig ( r) bracketed expressios. Hmmmmm...this choosig remids me of somethig... We have to choose r places from the possible positios (i a word of legth ) i which to put the a s ad fill the remaider with b s. There are ( r) ways of doig this. Thus, the coefficiet of a r b r is ( r. This ca also be writte as r) if we prefer. Q]... Prove that the sum of all the etries i the -th row of Pascal s triagle is 2. Hit, let a b i the Biomial theorem. The Biomial Theorem says (a + b) 0 a b

7 Settig a b gives or 2 0 ( + )