Putnam Training Exercise Counting, Probability, Pigeonhole Principle (Answers)

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1 Putam Traiig Exercise Coutig, Probability, Pigeohole Pricile (Aswers) November 24th, Fid the umber of iteger o-egative solutios to the followig Diohatie equatio: x 1 + x 2 + x 3 + x 4 + x 5 = Aswer: Rereset each o-egative iteger as a strig of strokes, e.g. 5 =. The there is a solutio to the give equatio for each ossible way to arrage 17 strokes ad 4 lus sigs, e.g. (x 1, x 2, x 3, x 4, x 5 ) = (2, 0, 3, 5, 7) would be So the umber of solutios coicides with the umber of ermutatios of 17 strokes ad 4 lus sigs, which is ( ) ( = 17+4 ) = Prove that the umber of subsets of {1, 2,..., } with odd cardiality is equal to the umber of subsets of eve cardiality. - Aswer: - First Solutio: The umber of subsets of {1, 2,..., } with odd cardiality is ( ) ( ) ( ) The umber of subsets of eve cardiality is cardiality is ( ) ( ) ( ) The differece is ± 3 = (1 1) = 0. - Secod Solutio: We defie a bijectio betwee the subsets with odd cardiality ad those with eve cardiality i the followig way: if S is a subset with a odd umber of elemets we ma it to S = S {1} if 1 S, or S = S \ {1} if 1 S.

2 3. Let a 1, a 2,..., a a ordered sequece of distict objects. A deragemet of this sequece is a ermutatio that leaves o object i its origial lace. For examle, if the origial sequece is {1, 2, 3, 4}, the {2, 4, 3, 1} is ot a deragemet, but {2, 1, 4, 3} is. Let be the robability that a radom ermutatio of elemets (with all ermutatios havig the same robability) turs out to be a deragemet. Fid lim. - Aswer: Let D deote the umber of deragemets of a -elemet sequece. We will rove that ( D = 1 1 1! + 1 2! + 1 ) ( 1). I fact, let s deote P i the set of ermutatios fixig elemet a i. The set of oderagemets are the elemets of the uio P 1 P 2 P, ad its umber ca be foud usig the iclusio-exclusio ricile: P 1 P 2 P = P i P i P j + P i P j P k i i j i j k i Each term of that exressio is the umber of ermutatios fixig a certai umber of elemets. The umber of ermutatios that fix m give elemets is ( m)!, ad sice there are ( ( m) ways of ickig those m elemets, the corresodig sum is ) m ( m)! =. Addig ad subtractig from the total umber of ermutatios m!, we get D = 1! + ( + ( 1) 2! = 1 1 1! + 1 2! + 1 ) ( 1). Next, we have = D = ! 2! ( 1) 1, ad the limit is lim ( 1) = = 1 e. =0 4. Peter tosses 25 fair cois ad Joh tosses 20 fair cois. What is the robability that they get the same umber of heads? - Aswer: The robability of Joh gettig heads is the same as gettig tails. So the roblem is equivalet to askig the robability of Joh gettig as may tails as the umber of heads gotte by Peter. So if Peter gets k heads ad Joh gets k tails, the Joh gets 20 k heads, ad that is the same as both gettig joitly a total of k + 20 k = 20 heads. So the robability asked is the same as that of gettig 20

3 heads after tossig = 45 cois, i.e.: ( ) (That is ) 5. From where he stads, oe ste toward the cliff would sed a druke ma over the edge of a cliff. He takes radom stes, either toward or away from the cliff. At ay ste his robability of takig a ste away is, of a ste toward the cliff 1. Fid his chace of escaig the cliff as a fuctio of. - Aswer: Let x be the distace from the ma to the edge measured i stes. For > 0, let P the robability that the druke ma eds u over the edge whe he starts at x = stes from the cliff. The P 1 = (1 ) + P 2. We ow rewrite P 2 i the followig way. Paths from x = 2 to x = 0 ca be broke ito two arts: a ath that goes from x = 2 to x = 1 for the first time, ad a ath that goes from x = 1 to x = 0. The robability of the latter is P 1, because the situatio is exactly the same as at the begiig. The robability of the former is also P 1, because the structure of roblem is idetical to the origial oe with x icreased by 1. Sice both robabilities are ideedet, we have P 2 = P 2 1. Hece P 1 = (1 ) + P 2 1. Solvig this equatio we get two solutios, amely P 1 = 1 ad P 1 = 1. We ow eed to determie which solutio goes with each value of. For = 1/2 both solutios agree, ad the P 1 = 1. For = 0 we have P 1 = 1, ad whe = 1, P 1 = 0, because the ma always walks away from the cliff. For 0 < < 1/2 the secod solutio is imossible, so we must have P 1 = 1. For 1/2 < 1 we have that the secod solutio is strictly less tha 1. By cotiuity P 1 caot take both values 1 ad 1 o the iterval (1/2, 1], so sice P 1 = 0 for = 1, we must have P 1 = 1 o that iterval. Hece, the robability of escaig the cliff is 0 if 0 1, 2 1 P 1 = 2 1 if 1 2 < We ick oits at radom o a circle. What is the robability that the ceter of the circle will be i the covex olygo with vertices at those oits? - Aswer: Label the oits x 0, x 1, x 2,..., x (x x 0.) The the ceter of the circle will ot be i the olygo if ad oly if oe of the arcs defied by two cosecutive oits (measured couterclockwise) is greater tha π. Let E k (k = 0,..., 1) be

4 the evet the arc from x k to the oit ext to x k (couterclockwise) is larger tha 1 π. The robability of each E k is obviously, because for it to hae all oits 2 1 other tha x k must lie i the same half-circle edig at x k. O the other had, the evets E 0, E 1,..., E 1 are icomatible, i.e., o two of them ca hae at the same time. The, the robability of oe of them haeig is the sum of the robabilities: P (E 0 or E 1 or or E 1 ) = P (E 0 ) + P (E 1 ) + + P (E 1 ) = Hece, the desired robability is Prove that ay ( + 1)-elemet subset of {1, 2,..., 2} cotais two itegers that are relatively rime. - Aswer: We divide the set ito -classes {1, 2}, {3, 4},..., {2 1, 2}. By the igeohole ricile, give + 1 elemets, at least two of them will be i the same class, {2k 1, 2k} (1 k ). But 2k 1 ad 2k are relatively rime because their differece is Prove that if we select + 1 umbers from the set S = {1, 2, 3,..., 2}, amog the umbers selected there are two such that oe is a multile of the other oe. - Aswer: For each odd umber α = 2k 1, k = 1,...,, let C α be the set of elemets x i S such that x = 2 i α for some i. The sets C 1, C 3,..., C 2 1 are a classificatio of S ito classes. By the igeohole ricile, give + 1 elemets of S, at least two of them will be i the same class. But ay two elemets of the same class C α verify that oe is a multile of the other oe. 9. Prove that amog five differet itegers there are always three with sum divisible by 3. - Aswer: Classify the umbers by their remider whe divided by 3. Either three of them will yield the same remider, ad their sum will be a multile of 3, or there will be at least a umber x r for each ossible remider r = 0, 1, 2, ad their sum x 0 + x 1 + x 2 will be a multile of 3 too. 10. Prove that every covex olyhedro has at least two faces with the same umber of edges. - Aswer: Let F be a face with the largest umber m of edges. The for the m + 1 faces cosistig of F ad its m eighbors the ossible umber of edges are 3, 4,..., m. These are oly m 2 ossibilities, hece the umber of edges must occur more tha oce.

5 11. I the Z Z iteger lattice made of all oits (x, y) with iteger coordiates (x, y Z), we ait all oits usig 2 differet colors, say read ad blue. Prove that the lattice cotais a rectagle with its four vertices aited with the same color. - Aswer: Cosider the 3 9 rectagle with oits of the form (i, j) with 0 i 9, 0 j 2, ad look at the sets of oits of the form {(i, 0), (i, 1), (i, 2)}. The ossible combiatios of colors of those oits are ( 3 3) = 8. But there are 9 such sets, so two of them, say {(i 1, 0), (i 1, 1), (i 1, 2)} ad {(i 2, 0), (i 2, 1), (i 2, 2)} (i 1 i 2 ) will have the same combiatio. Furthermore at least two of the oits i each set will have the same color, say (i 1, j 1 ) ad (i 1, j 2 ) (j 1 j 2 ), ad also (i 2, j 1 ) ad (i 2, j 2 ). The the rectagle with vertices (i 1, j 1 ), (i 1, j 2 ), (i 2, j 1 ), (i 2, j 2 ), will have those vertices aited with the same color. 12. (IMO 1972.) Prove that from te distict two-digit umbers, oe ca always choose two disjoit oemty subsets, so that their elemets have the same sum. - Aswer: A set of 10 elemets has = 1023 o-emty subsets. The ossible sums of at most te two-digit umbers caot be larger tha = 990. There are more subsets tha ossible sums, so two differet subsets S 1 ad S 2 must have the same sum. If S 1 S 2 = the we are doe. Otherwise remove the commo elemets ad we get two o-itersectig subsets with the same sum. 13. Let A be the set of all 8-digit umbers i base 3 (so they are writte with the digits 0,1,2 oly), icludig those with leadig zeroes such as Prove that give 4 elemets from A, two of them must coicide i at least 2 laces. - Aswer: For k = 1, 2,..., 8, look at the digit used i lace k for each of the 4 give elemets. Sice there are oly 3 available digits, two of the elemets will use the same digit i lace k, so they coicide at that lace. Hece at each lace, there are at least two elemets that coicide at that lace. Pick ay air of such elemets for each of the 8 laces. Sice there are 8 laces we will have 8 airs of elemets, but there are oly ( 4 2) = 6 two-elemet subsets i a 4-elemet set, so two of the airs will be the same air, ad the elemets of that air will coicide i two differet laces.