Question

(a) Suppose that A1,..., All is a collection of k > 2 sets. Show that U412 4:1 - L140.4jl. i=1 {ij} where the second term on the right sums over all subsets of [k] of size 2. [Hint: Use induction on k] (b) Deduce that in every collection of 5 subsets of size 6 drawn from {1,2,..., 15}, at least two of the subsets must intersect in at least two points. (c) Show that the inequality in (a) is an equality if there is no element that lies in three of the sets in the collection. (Hint: Use the first form of the P.I.E.]

Answer 1

to for two sets A, and A₃ we have JAUA2I=JA, 1+1A21-ANALI this the inequility is true for a k=2. for k=3 consider 3 sets A., A2, Az we know that {ij} Unil=:- Itines 1 + ANAMAI - 18 A = {1:1-SIANASI Let the inequility is towe for k=m.de w isi w w LE ElAil-ElAin Ail is 11 Now for k=m+1 consider A, A2,..., Am, Amt Sets. het B = UAi then IBUAmA = 1B1+1 Antil-1 Bn Amell [Since this is true > ΣΑι -ΣΑ ΔΑ; for k=2 ]. i {i, i} Thus By mathematical induction the result is true for all KEN. by Ler Ai S {1,2, . -., 15} for i= 1, 2, 3, 4, 5 and Ail=6 for i=1,2,3,4,5 . vory from previous inequitity, IVAL = 1 Al-Elainas ist st Sis? ZADA; 2 30-15= 15. = 157, 30-5. JANA; Nowy from the set {1, 2, 3, 4, 5} two numbers &2,3} can be select in 5c - 10 ways it each Ain Aj contains one element then ElAnAil will be 10 & 15. $1,51 So, at least two sets most intersect in at least two points.

we know poin care's formular TOATE ŠInil- È Innalt +(-1*3!ANAL, D--NA, asicisn i s iciz< <ipan Son of there is no elements lies in three sets, then for any s iz t 8 1,2,...,n} AnAin Aiz=& in AinAin Aizl=0 there fare for an I AinAin.nAinl=0. 154 cizconcupen Hence lů Al- ŽIAI-E TANA; Henca in 1cicisn.