germanandres82 wrote:
I am honestly completely lost on this one, seems that I have a completely different idea on what the problem is asking.
When I read "cardinality", I understand basically the number of elements in a set. Following this, my interpretation of the question is if I can know the number of elements in set A.
1) 2 is the cardinality of exactly 6 subsets of set A.
There are 6 subsets in set A with 2 elements INSUFFICIENT - THERE COULD BE OTHER SUBSETS IN A OR EVEN SOME OF THE ELEMENTS CAN BELONG TO MULTIPLE SUBSETS
II) Set A has a total of 16 subsets, including the empty set and set A itself INSUFFICIENT - NO INFORMATION ON NUMBER OF ELEMENTS IN SET A
I)-II) INSUFFICIENT - WE KNOW THAT 6 OUT OF THE 16 SUBSETS OF SET A HAVE " ELEMENTS; YET WE DO NOT KNOW HOW MANY ELEMENTS THE OTHER 10 SUBSETS HAVE NOR IF SOME ELEMENTS ARE IN MORE THAN 1 SUBSET
Clearly the question is going in a completely different direction than my line of though, can someone PLEASE help me explain where do I get completely lost?
Thank you!!!
germanandres82When I read "cardinality", I understand basically the number of elements in a set. Following this, my interpretation of the question is if I can know the number of elements in set A. --> Yes, you're understanding of cardinality is fine
There are 6 subsets in set A with 2 elements INSUFFICIENT - THERE COULD BE OTHER SUBSETS IN A OR EVEN SOME OF THE ELEMENTS CAN BELONG TO MULTIPLE SUBSETS --> Not true!!
Can you give any example to prove ??
I will try and explain.
Case 1: Let \(A_1\) has 1 element in it
Assume Set \(A_1\) = {1}
Number of subsets with '0' elements possible = {} --> 1
Number of subsets with '1' element possible = {1} --> 1
--> Total number of subsets possible for \(A_1\) = {}, {1} = 2 or \(2^1\)
Case 2: Let \(A_2\) has 2 elements in it
Assume Set \(A_2\) = {1, 2}
Number of subsets with '0' elements possible = {} --> 1
Number of subsets with '1' element possible = {1}, {2} --> 2
Number of subsets with '2' elements possible = {1, 2} --> 1
--> Total number of subsets possible for \(A_2\) = {}, {1}, {2}, {1, 2} = 4 or \(2^2\)
Case 3: Let \(A_3\) has 3 elements in it
Assume Set \(A_3\) = {1, 2, 3}
Number of subsets with '0' elements possible = {} --> 1
Number of subsets with '1' element possible = {1}, {2}, {3} --> 3
Number of subsets with '2' elements possible = {1, 2}, {2, 3}, {3, 1} --> 3
Number of subsets with '3' elements possible = {1, 2, 3} --> 1
--> Total number of subsets possible for \(A_3\) = {}, {1}, {2}, {3}, {1, 2}, {2, 3}, {3, 1}, {1, 2, 3} = 1 + 3 + 3 + 1 = 8 or \(2^3\)
Case 4: Let \(A_4\) has 4 elements in it
Assume Set \(A_4\) = {1, 2, 3, 4}
Number of subsets with '0' elements possible = {} --> 1
Number of subsets with '1' element possible = {1}, {2}, {3}, {4} --> 4
Number of subsets with '2' elements possible =
{1, 2}, {2, 3}, {3, 4}, {4, 1}, {1, 3}, {2, 4} --> 6
Number of subsets with '3' elements possible = {1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {2, 1, 4} --> 4
Number of subsets with '4' elements possible = {1, 2, 3, 4} --> 4
--> Total number of subsets possible for \(A_4\) = {}, {1}, {2}, {3}, {4}, {1, 2}, {2, 3}, {3, 4}, {4, 1}, {1, 3}, {2, 4}, {1, 2, 3}, {1, 3, 4}, {2, 3, 4}, {2, 1, 4}, {1, 2, 3, 4}
= 1 + 4 + 6 + 4 + 1 = 16 or \(2^4\)
NOW THIS IS EXACTLY WHAT WAS GIVEN IN STATEMENT (1), "
2 is the cardinality of exactly 6 subsets of set A" = "
{1, 2}, {2, 3}, {3, 4}, {4, 1}, {1, 3}, {2, 4} --> 6"
--> It meant exactly 6 subsets of A has cardinality 2. So, We can DEFINITELY say that Cardinality (number of elements) of set A = 4 as per Case 4 -->
SufficientNOW, Lets consider statement 2
Set A has a total of 16 subsets, including the empty set and set A itself INSUFFICIENT - NO INFORMATION ON NUMBER OF ELEMENTS IN SET A
As per case 4, which has 16 subsets, we can DEFINITELY say that number of elements of set A = 4 (cardinality) -->
SufficientSo, answer is D
Hope it's clear!