sect is defined as collection of well-defined objects the meaning of well-defined objects is let us take one example two four six eight ten all these are called as even numbers so these are the numbers of well-defined property called even numbers take one more example one three five seven nine these are all the well-defined numbers of odd number property so this is called as a set is equal to set off the set will be denoted by the symbol flower brackets now in this set we have one two three four numbers but in sets chapter we’ll call these numbers as elements so how many elements are they one two three four so number of elements present in a set these for now take like a second example B is equal to sit off five six seven eight nine so these are all called the elements number of elements present in B set is total fine so over all the elements which are present in that particular set we can count it as number of elements we have three types of sets finite sit infinite sit and empty set finite sit means you should able to identify first element and last element in this set is equal to set of one two three four we have four elements in this first element is one last element is four we are able to identify both first and last so this is called finite set let us take second one as infinite set take B is equal to set of 5 6 7 8 9 so on in this we are able to identify only the first element that is fine we are unable to identify the last element so this is called infinite set third one is empty set in empty set we don’t have any elements then that set is called empty set now we are going to start operations on sets so first operation is Union Union is nothing but combination of both the elements of sets let us take the example a is equal to set off 1 2 3 B is equal to set of 4 5 6 a Union is the vert we will indicate the symbol by the letter U so a union B is equal to a set is 1 2 3 again Union word will denote with the symbol U 4 5 6 Union means combination so first what elements are there in the asset as it is will write 1 2 3 next we say telemon’s are 4 5 6 so let us continue 4 5 6 so a union B is a set of combination of a set elements and B settlements now second one is intersection the elements which are present in both the sets let us take example a is equal to 1 2 3 4 B is equal to 3 4 5 6 both means here 3 is present in asset and 3 is present in visit so we lit in a intersection B 3 and here 4 is present in asset and 4 is present in B sit so a intersection B is the elements which are present in both the next one is difference the elements which are present in only in the first set a is equal to 1 2 3 4 5 B is equal to 4 5 6 so difference is the symbol will denote with the later my wrote with the symbol minus 1 2 3 4 5 4 5 6 7 8 4 4 will get cancelled because they present in both 5 5 will get cancelled so the elements which are remaining in only the first set 1 2 3 so in difference a minus B the elements are 1 2 3 No the properties of sets first one is Union a union B is equal to B Union a let us take example a is equal to set off ABCDE B is equal to say Talk CDF

first a union B a set elements Union is denoted with the letter U visa elements I already said Union means is the combination of both the sets so first right which set is area a so write all the a settlements a b c d e and next b settlements C D already we have written here so repetitions are not allowed so Balan left out element is yes so in union B is ABCDE F next one B Union a we set elements are C D F a settlements are ABCD here also B is the first set so first threat all the elements of b c d e f a set elements a b already we have written CD repetitions are not allowed when we observe first equation and second equation both are equal so here is the starting element c is the starting element but even though the total number of elements of a union B is 6 B Union a is also 6 so therefore our Union property is verified a union B is equal to B Union a this is also called committed to property of Union let us see in the Venn diagram and B are the two sets when we represent in the circles two circles will intersect this is called intersection area in a and B CDER the intersection elements so we are written in the intersection path in acid the left out elements are a comma B so it will write it in the a B and in B set F is a element so we need to write it here so when we are asking to shade a union B you need to share completely whole set now similarly B and a also first write the common element CD enter in the intersection area in B said the balanced element is F in a set the balance and went is a B so we need to shade when we observe and diagram also a union B is also same Union is also same so this is called Union committed to property now second property’s intersection committed to off intersection intersection B is equal to B intersection a so let us take example 1 2 3 4 in is it 2 3 4 5 in B set intersection B as I already said intersection is nothing but common elements in a set 1 2 3 4 is common in B seat also 2 3 4 is common so in a intersection B 2 3 4 is the elements now B intersection a in be set also 5 2 3 4 and in a set 1 2 3 4 in these two sets the common elements are 2 3 4 therefore a intersection B is equal to B intersection a which is a commutative property of intersection let us see the Venn diagram of a intersection B so this is the called intersection region so a intersection B is already known as common 2 3 4 so let us enter the values of 2 3 4 in the intersection region and in a set the balanced element is 1 so we need to enter here and in B set the balanced element is 5 so we need to enter here since it is intersection right diagram we need to shared only the intersection part similarly B intersection a the intersection elements are 2 3 4 in B set 5 is the element here and in a set the balanced almighty’s 1 since it is intersection we need to shade only the intersection region so this is called the intersection committed to property you properties associative property in associative we need three sets a B and C so in associative first properties Union a union B Union C is equal to a union B Union C when we are pronouncing also a union B is in the brackets so we need to read as a union B Union C is equal to a union B Union C now let us take example a equal to 1 2 3 B equal to 2 3 4 C equal to 3 4 5 so first a union B Union C a union B you need to combine all the elements of a and B repetitions are not allowed is it 1 2 3 in we said 2 3 already we have entered so balance element for union C set is 3 4 5 so Union is a combination of both the sets so first a union B elements will write it in answer 1 2 3 4 already 3 & 4 we entered here so the left out element is 5 so this is first

equation now a union B Union C so a set is 1 2 3 Union of B Union C let us see B is 8 is 2 3 4 so 2 3 4 in C’s it also 3 4 already we have written here so since repetitions are not allowed so C set balance element is fine again we need to combine these two set elements 1 2 3 4 5 let it be second equation hence first and second equations are equal our associative property of Union is verified so now let us see the associative property through n diagram since in associative we have three sets a B and C circles will be there and 3 3 circles will have common intersection region is here so in all the 3 sets we find 3 is the common element so we have written 3 common element in the intersection part of all the 3 circles here now a and B in a and B 2 is a common element so we have written here in B and C B and C we have the common element called 4 we have returned here so the left out element of a set is 1 so we have return here since we set elements all the three are accompanied here so we no need to write any element in the pset now see set balanced element is five we need to enter here since it is Union we need to share all the portions of the Venn diagram similarly a union of B Union C the entering the values of both the Venn diagrams are same so once again since all the three set of having common element three will be entered here a and B common element is 2 B and C common element is four since wheel element does not have any element no need to enter anything here in asset left out element is 1 so we have entered here the left out element is fine since it is Union we need to shade all the portions of the Venn diagram when we compare the answer and the Venn diagram both are verified hence this is the associative property of Union so the next property of associative is intersection a intersection B intersection C equal to a intersection B intersection C let us take example is equal to ABC elements B is equal to B CD C is equal to C D so a intersection B intersection C in a and B B and C are the elements common so we have written here a intersection B means B comma C intersection C set elements are C D now in this set and this set C is the only element common so we have written C let it be equation 1 now take a intersection B intersection C is it is ABC elements intersection in B and C C comma D are common now in these two C’s the element is common let it be equation 2 when we compare both the equations both the equations are having the same settlement C therefore associative property of intersection is verified so let us see the associative property of intersection through Venn diagram since we know already no associative property will have 3 sets a B and C now in all the 3 sets C is a common element so this is the area which we have to write the all the three sets common elements in C is the element have written here C now a and B B comma C is a common element already have entered see here so B is the only element I can enter in the region a intersection B B intersection C also see comedy are the common elements but always in C I have entered here in B intersection C common element is d in a and C said we don’t have any common elements so no element will be entered here so we need to find which is a common element in all the 3 is C so we need to share this region here similarly the other side intersection of B intersection C entry in the Venn diagram is same so in all the three set element C is the common and B common element is B B and C common element is d since we have associative property of intersection we need to share the region of see only hence both the intersection property of associative is verified through calculation and Venn diagram the next properties distributive property in this also we have two types Union over intersection in exam instead of learning the property learn according to the words Union over intersection since Union is the first word a union over indicates brackets intersection

means B intersection C is equal to when we are learning the property multiply this a Union to be a union into B is a union B next intersection symbol here a Union into C is a union C so the distributive property Union over intersection Union word is a first means a Union over means brackets B intersection C is equal to a union into B’s a union B intersection a union into C’s a union C let us take the example k equal to 3 5 7 9 L equal to 5 8 9 m equal to 1 2 3 9 now let us take instead of ABC we have used the latest klm so K Union over means brackets L intersection M first the brackets part we need to solve L intersection M intersection means common element in 5 8 9 and 1 2 3 9 9 is a common element so in L intersection M place we need to write the element 9 case it is 3 5 7 9 so Union means all the elements so 3 5 7 I let it be equation 1 now we need to prove K Union L intersection K Union M K and L sets we need to combine Union means all the elements so k Union L values are 3 5 7 8 9 K Union M is 1 2 3 5 7 9 intersection means we need to write only the common elements 3 5 7 9 let it be equation 2 laterz now see compare first equation and second equation since both are equal Union over intersection property is verified through calculation now let us see the Venn diagram now since it is distributive property we use three sets klm these are the three circles and this is the intersection part of all the three circles in klm sets we have nine is the common element so we need to enter nine year K&L sets common element is five K and M common element is three in L and M we don’t have any common elements so we need not write anything in case it we have total elements three five seven nine three five nine we already entered here the left out element is seven same way in L set we have the elements five nine three seven since already entered here eight is the balanced element we need to enter here and in M set the left out element is one two since our property is K Union L intersection iam first of all L and M intersection region is this so we need to shade first this then K Union completely K circle so now our Venn diagram represent K completely circled and L intersection iam so this is the shaded portion of K Union of L intersection M same way this is the Venn diagram of K Union L intersection K Union M since therefore both the Venn diagrams are equal hence our distributive property Union our intersection is verified through Venn diagram and calculation so in distributive second type of properties intersection over Union in exam if they give intersection over Union a first word is intersection so intersection symbol over indicates brackets B Union C intersection B how to remember is a intersection into B is a intersection B Union a intersection into C is a intersection C let us take the example k equal to set of XY Z comma T L is equal to Y and Jack m equal to rst now k intersection L Union M case the Tacitus will write x comma Y comma Z comma T intersection L and M since Union is the symbol we need to write all the elements so Y jet rst when we are finally finding the value K intersection this value intersection is nothing but common in these two sets we have the common elements Y comma Z comma T let it be equation 1 now k intersection yell in K and L we have the common elements Y comma J in K intersection M in K and then we have the common element T again we need to come and the these both the values with the operation Union Union indicates all or combination of both the elements so why come as yet come at a let it be equation 2 when we see first in second equations both the answers are same

hence intersection over Union properties verified now let us see this example through Venn diagram since this is distributive property intersection over Union we use three sets okay Ln M so when we observe K and L common elements are y comma J so this is the region where we need to write k intersection L and K intersection M common element is T hence we don’t have any common element in all the three sets no value should be entered here this is the region where we need to enter L and M common elements since we don’t have any common we don’t we no need to enter any value so k intersection of K Union L intersection K Union iam when diagram is this K&L common elements are why and jet so we need to enter in this region K and M common element is T so we need to enter in this region since we need to combine both these Venn diagrams will get K u intersection a union M is this is the shaded region with both the Venn diagrams our scene so therefore distributive property intersection over Union is verified through calculation as well as Venn diagram hence distributive property is verified that an extra party is de Morgan’s laws this is the first law a union B whole complement whole indicates brackets complement is equal to a into complement is a complement Union opposite is intersection being to complement this B complement you need to remember the formula in such a way a union B whole complement equal to in the complement is a complement Union opposite this intersection being the complement is B complement let us verify this property by using example equal to 2 comma 4 be equal to 3 4 5 and you is a universal set this is the main set these two are the subsets we can write from this set that subset this these are the numbers present in the universal set 1 2 3 4 5 now let us write the de Morgan’s love first love first we need to prove a union B whole complement complement indicates you – of whatever is there in the brackets we need to write you – of a union B you said this 1 2 3 4 5 – a union B is in a and B we need to read all the elements because you indicates union so we combined both the elements will get 2 3 4 5 I already explained you about the difference first we need to cancel the common elements and we need to write the balance element in the first set since 1 is element is balanced here we will write let it be equation 1 now we need to calculate a complement whenever we see complement we need to write you – whatever said they are given here a know you set is 1 2 3 4 5 – a set is 2 3 4 cancel the common elements 2 3 4 Heaney acid the balanced elements are 1 & 5 same way we need to calculate B complement complement means you – since B is the set here you will write you minus B you set is 1 2 3 4 5 – we set is 3 4 5 both will get canceled 1 comma 2 a complement intersection B complement a complement value we got 1 comma 5 B complement value 1 comma 2 a complement intersection B complement in these two which element is common one is the common so we have written here 1 let it be equation 2 now let us compare equation 1 and equation 2 both the answers are same so therefore de Morgan’s law first lies verified through calculation so now let us see this de Morgan’s first law through Venn diagram first we need to calculate a union B Venn diagram a and B are two sets so we have drawn two circles me and B common element is 4 we need to write here and in essence the left out element is 2 and then we said the left hotel went is 3 & 5 U is a universal set will write it in outside the circle will denote with the letter U in that the balance element is 1 and a union B all complement we need to say shade the region which is outer region not the two circles outer region where Universal set one element will lies now a complement the entering the values is same if it is a complement other than the a circle rest all the parts of the circle should be shaded including the B if it is B complement B circle only we will not say rest all we will shade a union B a intersection B complement is the outer region only we need to shade where you universal set realize the element 1 since we are de Morgan’s laws when diagram is verified

therefore de Morgan’s first lies verified through calculation and when diagram therefore the Morgan’s first lies verified so now let us see DeMorgan second law intersection behold complement equal to a complement Union B club B complement here also will read the formula ain’t the compliment is a complement intersection opposite is Union we come in to complement is B complement let us throw this law according to this example you is a universal set 1 2 3 4 5 equal to 2 4 be equal to 3 4 5 so now let us prove a intersection B hold complement complement indicates you minus whatever is there given the said the same thing a intersection B in a and before is the common element so we need to write in a intersection B place 4 4 4 will get cancelled the left out elements in the first set are 1 2 3 5 let it be equation 1 now find a complement complement means you – given satis y8 so use it – is it in yes at – 4 is they cancel 2 4 in use it the balance elements 1 3 5 have been written here same way B complement B complement is you minus B you set is 1 2 3 4 5 – 3 4 5 cancel 3 4 5 common elements here 1 comma 2 is left out here now a complement Union B complement in these two since Union symbol is there we need to combine both the values 1 is present in both so since repetitions are not allowed only one time we need to write 2 3 5 let it be equation 2 if you observe equation 1 and equation 2 both are same so ends our D Morgan second law is verified through calculation so now let us verify the de Morgan’s a kenda through Venn diagram first a intersection B in a certain B said the common element is 4 we have written here in a set the left out element is 2 in be set the left out elements are three and five Universal set will right out of the sets where you will indicate so 1 is the element common now compliment when diagram other than the for the rest all the portion should be shaded so this is the diagram of a intersection B whole complement same way we need to calculate a complement one diagram a complement this other than the a circle rest of the all the portion should be shaded B complement is nothing but except the B circle rest all should be shaded since it is Union we need to combine both the diagrams we will get the Venn diagram with only intersection region will not be shaded rest all will be shaded hence de Morgan’s now is verified through when diagram anding Morgan secondly is verified from first and second equations therefore sakti Morgan’s second lie is verified you so last in topic is disjoint sets and non disjoint sets disjoint sets is nothing but enough a intersection B if you don’t have any common elements then that sets are called disjoint elements so once again disjoint sets are nothing but if you don’t have any common elements then it is called disjoint sets and number of elements in a intersection B is 0 so when we don’t have elements we don’t have any number of elements so it is 0 so the formula is n of Union B number of elements of a union B is equal to ad number of elements of a plus number of elements of B both the size the values will be same in case if it is non disjoint sets not disjoint sets means intersection B will have common elements so the formula will be n of a union B nothing but number of elements present in a union B equal to n of a number of elements of a plus number of elements of B minus n of a intro intersection B you so far we discussed the properties of sex let us start the exercise this is the first problem is equal to set of lmnopq basic melt is a tough ml o rst we need to verify the commutative property of Union and intersection so let us

start with Union a union B this is the a set elements and this is the base of tenements you already know that your in means combination of both the sets so here which is the set is that yes it so first complete there is at elements l m n o p q then you write the balanced elements of B since repetition is not allowed same way B Union a so first write the B set elements MN o rst then is a Tillman’s let it be equation 1 this is and this is equation 2 since both are equal commutative property of Union is verified same way so now we are going to prove the intersection committed to property a intersection B value we need to calculate a is a set lmnopq B is the set MNO rst we already know intersection means common elements here M is there here is M is also there so M we need to write and O so which are the common elements are MN go in a intersection B say may be calculate B intersection here we are going to get em men go hence a intersection B equal to B intersection a committed the property of intersection is verified so now see the second problem Pisa set ABCD are the elements Q is the set a II I’ve oh you are is a set a CH g we nee are going to verify the associative property of Union and intersection now let us start with union B Union Q Union are first we need to calculate which is given in the brackets P Union Q so Union eans combination of both the sets so a b c d e since a and d already we have written here a part of other that why were you so your P Union Q elements are a b c d e i o u union are settlements are a c e g again we need to combine these both settlements so ABCDE i go Yugi so let it be equation one in Union repetition is not allowed same way the union of Q Union are first the brackets one should be solved Q Union are you set is a e I go you and I and II already we have written here so we are going to write C and G extra now P Union this one is combine all the elements a b c d e g io you if you count the elements of this set a on line and this set is also 9 therefore associative property of union is verified now let us see the case of associative property of intersection B intersection Q intersection are here also we need to solve first the brackets 1 P and Q now see in your problem in P and Q which are the elements common is present here is also present here is present in q sit as well as piece it so intersection you means common elements so a comma E again intersection with asset a see eg now again in these two we need to copy check which are the common elements a a en e so final answer is a comma e same way P intersection Q intersection up in Q and are also we have any common elements again we need to combine this set with piece it again we are get in the common elements a comma I let it be equation 2 when we compare equation 1 and 2 we got the common elements in the both the sets as a comma E and a kawaii so associative property of intersection is also verified so now our third problem is verifying de Morgan’s now this is a de Morgan’s first law they gave use it is it and B sit first you need to calculate a union B whole complement as I said complement is nothing but you – hey Union B use it already they are given same thing as this will right – a union B a and B we need to come back so we are return 4 8 16 24 20 28 first cancel the common elements common ways it should present in both 4 is there here in both so we have cancelled 8 16 20 24 28 so the left out element is 12 so let it be equation 1 now a complement we need to calculate complement means you – a use it subtracted by asset subtraction is nothing but cancel the common elements and write the balance elements in the use it so balance elements are for 12 20

28 so we have written here so same way B complement so B complement balance elements are 8 12 24 so a complement intersection B complement in these two sets we have 12 is the common element so final answer is 12 let it be equation to compare equation 2 by equation 1 so in de Morgan’s law both equation 1 and equation 2 are equal therefore your de Morgan’s first lies verified you so now see the next X is his first main first question and no phase number of elements in a set is 37 and off B means number of elements present in B satisfy 6n of a union B is number of elements present in a union B 51 so we need to calculate number of elements present in a intersection B we already studied the formula disjoint sets formula n of a union B equal to n of a plus n of B minus n of a intersection B the values of this formula already they were given so now in place of n of a union B we are going to substitute the value 51 in n off a 37 in off be 26 and we need to calculate this n of a intersection B so minus values when it comes to opposite side will become plus 37 plus 26 will become 63 plus 51 when it goes to opposite side it will become minus 51 63 minus 51 is 12 therefore n of a intersection B value is 12 so here we have total number of persons as 50 and of T is number of people drinks T is 30 an Aussie number of people drinks coffee is 25 n of the intersection C number of persons will drink both 16 we need to calculate a dirty or coffee second one is neither tea nor coffee so either tea or coffee is nothing but Union value we need to calculate so according to disjoint sets formula n of T Union C equal to n of t plus n of C minus n of T intersection C n of T value is 30 anak C value is 25 n of T intersection C value is 16 30 plus 25 is 55 55 minus 16 is 39 let us see this one through n diagram now let us see this 39 when how to represent in the Venn diagram first of all T and C how many people are drinking 16 so you need to enter 16 value in the intersection region and only T how many are drinking means enough T they gave 30 subtract 30 of Z out of 16 so 30 minus 16 is 14 same way coffee 25 are drinking among them only coffee or many are drinking 25 minus 16 9 when you add all these three 14 plus 16 plus 9 we are getting 39 so either tea or coffee how many persons are drinking is 39 now second one we need to calculate neither tea nor coffee how many people have this there 50 so 50 minus 8 of tea or coffee is ringing by 39 percent so 50 – 39 11 so 11 people will not drink neither tea nor coffee you in a survey of 150 people the observed number of people are having eye problem is ninety number of people having a heart problem is fifty and both eye and heart problems are thirty we need to calculate how many are having either I or heart problem so according to the disjoint sets property enough Union H is equal to n of e + n office – n of e intersection H n of e value is 90 n of H value is 50 and of intersection H is 30 90 plus 50 is 140 140 minus 30 is 110 now let us check whether this value we will get through n diagram now see here we have two sets if an IH for heart both of the problems having number of people is 30 so we need to write it in the intersection region I problem they give

90 but how many are having only eye problem 90 minus 30 is 60 same way heart problem is given by 50 only heart problem how many are having 50 minus 30 is 20 I had all this we are getting 110 so our calculation for either IRR problem is same now they have asked the question percentage of people either I union heart so we got total people are 150 and we got the value either heart or I problem is 110 so 110 by 150 into 100 Y under we are multiplying because we need to calculate percentage 0 0 will get cancelled 11 in 200 is 1100 divided by 15 will get seventy three point three three percent you so let us take total number of farmers in a village is 120 number of farmers will grow vegetables 93 flower sixty-three sugarcane 45 vegetables and flowers 45 flowers and sugarcane 24 sugarcane and vegetables are 20 now we need to calculate how many will grow all the three so far love how many will grow all the three is total farmers are 120 – enough we plus an office F plus an office all the three we need to add and subtract it from enough B intersection yes F intersection yes we intersection years so 93 plus 63 plus 45 is 2 not 1 45 plus 24 plus 27 is 96 when you subtract these two a good thing or not 5 120 minus 1 not 5 is 15 so let us represent this given data through n diagram we got 15 former so right 15 farmers in the intersection region now we intersection your face 45 already 15 are there years so that they will be the balance that they plus 2 think 45 we leave it same way F intersection yes F intersection already 15 are years so 24 minus 15 is 9 add these two you are getting 15 plus 9 24 we intersection yes we have 27 now given 15 is already entered year so 12 12 plus 15 27 if you want to know how many grow only vegetables add all the three subtracted from 93 30 plus 15 45 45 plus 12 57 so 93 minus 57 is 36 same way for sugarcane also 12 plus 15 27 27 plus 9 36 so 36 45 minus 36 is 9 same wave of flowers 30 plus 15 45 45 plus 9 50 54 so 63 minus 54 is 9 so this is the way how to represent the total number of farmers brought all the 3 this is 120 is a total for us who are not five we have calculated 120 minus one not five is 15 farmers are growing all the three vegetables flowers and sugarcane so now recap solution what and Oliver studied in assets chapter first one is kama tative a union B equal to B Union a which is in union a intersection B equal to B intersection a in intersection next associative property Union and intersection distributive property Union / intersection intersection over Union and de Morgan’s laws first love then second love now last one is disjoint sets formula and of Indian be equal to n of a plus n of B minus n of a intersection B