symmetric difference

Question

If A and B are subsets  of set X={1,2,3........100} and A∆B denotes set of all elements of X which belong to exactly one of A or B.Then total no of subsets of X such that A∆B ={2,4,6.....100}is

a)2^⁵⁰

2)2^⁵¹

3)2^¹⁰⁰

4)2^²⁵

Answer 1

2^100. Each of the odd elements of X have 2 choices, either they can be in both A & B or they can't be in both A & B, so 2^50 choices for them. Each of the even elements of the set can either be in A or be in B, so 2^50 choices for them. Thus overall 2^100 choices

Answer 2

According to the question  A $\Delta$ B should contain all even number of elements which in set notation can be written as :

(A $-$ B ) $\bigcup$ (B $-$ A ) = {2,4,6,........100}

No of ways 50 elements can be either in A or in B is 2$^{50}$ .

Now Both set A and B either  should contain any of the odd elements {1,3,5,......99} or with will not contain any of the odd elements ie it has each of 50 odd elements have two chances either to be placed in both A and B or to be placed outside of both A and B  which can be done in 2$^{50}$ ways .

Hence the total number of subsets is : 2$^{50}$ * 2$^{50}$ = 2$^{100}$.

Answer option C .

Plz correct me if am getting wrong .

Comments

Let me know if my explanation is ok with you and also whether you find it logical or not .

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Nice logic , just one confusion in ur explanation :

u chose to keep the elements both in A and B , i.e.in intersection portion of A and B ,Now if I have outside it i.e. in complement of A UNION B , then what will we be the problem since then also we will have the symmetric difference to have to consist of even numbers only 

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When did i say there will be problem ? Since i can place the element(odd numbers) either in their intersection or outside their union that why i have 2 choices for each element . You can choose to keep all outside those are all the parts includes is : 2$^{50}$ .

Answer 3

OPTION C is correct

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A∆B  = (A∪B) - (A∩B)

given that A∆B contains {2,4,6,---100)

case-1: only with even no's like         

  A                                B

{2,4,6,---------98,100}       {∅}    

{4,6,8,-------98,100}         {2}  -[we can choose one of  50 elements=50c_{1 ways}]

//y {6,8,--98,100}            {2,4} -[ there are 50c₂ possibilities)

like that totally

50c₀ + 50c₁ + 50c_{2 +----+} 50c₅₀ =2^50 sets exist.

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case 2:

now consider odd no's {1,3,5,------99)[ we can add odd no's to the both sets like]

  A                               B

{1,2,4,6,------,100}         {1}   -[we can choose one of 50 elements =50c_{1  ways})

{1,3,2,4,--,100}             {1,3}  -[ we can choose two elements=50c_{2  ways})

like that totally

50c₀ + 50c₁ + 50c_{2 +----+} 50c₅₀ = 2^50 sets exist.

( these odd elements we can add to above (case-1) all the sets)

:: total 2^50 *2^50 =2^100 exists.

// up vote if you agree