The Cardinality of a Finite Set is the Number of Elements in the Set.

Question: The cardinality of a finite set is the number of elements in the set. What is the cardinality of set A ?

(1) 2 is the cardinality of exactly 6 subsets of set A.
(2) Set A has a total of 16 subsets, including the empty set and set A itself.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.

This topic is a part of GMAT Quantitative reasoning section of GMAT. This question has been taken from the book "GMAT Prep Course" published in the year 2004. This is a GMAT Data Sufficiency question that allows the candidates to prove if the statement is sufficient as per the options. The total time allotted for this part is 62 minutes which allows the candidate 2 minutes to answer each question.

Solution and Explanation:

Approach Solution 1:

The problem statement is asking us - How many elements are there in Set A

We will be checking each statement one by one.

Statement 1: 2 is the cardinality of exactly 6 subsets of set A.

We will be simplifying the statement for understanding-
Statement 1 says that - 6 subsets of set A can be formed picking two elements

For example - If there is set A with 3 elements (1,2,3)- i can have subset with no element, second subset with 1 element each , third subset with 2 elements and fourth set of subset with 3 elements in it.

3C0+3C1+3C2+3C3= Total Subset of set A

So for the given set in the problem statement ,lets say it has n elements
Statement 1 gives us information - that nC2=6

We need to find n here.

To calculate n here lets solve
nC2= n!/2! (n-2)!
= n(n-1) (n-2)!/2! (n-2)!

Simplifying we can write n! as n (n-1) (n-2)! [ for understanding- if n=6 ,than 6! can be written as 6*5*4!--same way]

= n(n-1)/2! [(n-2)! cancels out]
=n(n-1)/2 [ we know 2! =2*1=2]

Statement 1 says nC2= 6

n(n-1)/2= 6

Solving further n(n-1)=12
n^2-n-12=0
n=4 or n=-3

obviously it cannot be -3 , so we know n=4.. Set A has 4 elements. Hence, statement 1 is sufficient.

Statement 2: Set A has a total of 16 subsets, including the empty set and set A itself.

Total subsets=16 (given)

We need to remember that - Number of subsets for a set of n elements= 2^n

We know 2^n=16
n=4

Hence, it is Sufficient.

Since each statement alone is sufficient, D is the correct answer.

Approach Solution 2:

Let 'n' be the cardinality of set A.

Statement 1:
2 is the cardinality of exactly 6 subsets of set A.
Number of 2 sets that can be formed from a set of n elements = nc2 = 6
--> n(n-1)/2 = 6
--> n(n-1) = 12 = 4*3
By comparison, n = 4
So, Cardinality of set A = 4 -- Sufficient

Formula: Number of subsets possible for a set of n elements =
2^n (including the empty set & the set itself)
E.g: If Set A = {1, 2, 3}, All subsets possible = {1}, {2}, {3}, {1,2}, {2,3}, {1,3}, {1,2,3}, {} = 8 subsets i.e, 2^3

Statement 2:
Set A has a total of 16 subsets, including the empty set and set A itself.
--> 2^n = 16
--> n = 4
So, Cardinality of set A = 4 -- Sufficient

Since both statements alone is sufficient, D is the correct answer.

Approach Solution 3:

The first statement says: "The number of 2 cardinality subsets in set A is exactly 6". This means that there are exactly 6 two-cardinality subsets in A (no more, no less).

Keeping the above fact in mind, we can analyze the two statements:

(1) If there are exactly 6 subsets which have cardinality = 2 in set A that means we can make exactly 6 groups of 2 elements out of set A.

Let n be the number of elements in set A. C(n¦2) = n(n-1)(n-2)!/(n-2)!(2!) = 6; from here we have n = 4.
Hence, it is Sufficient.

(2) Set A has exactly 16 subsets (no more, no less). That means we can make exactly 16 different combinations out of n.

#of combinations we can take out of any set A is: C(n¦0) + C(n¦1) +...+ C(n¦n) = 16.

Now we will see which n satisfies this condition?

Let's try with n = 1 --> C(1¦0) + C(1¦1) = 2. Nope.
Let's try with n = 2 --> C(2¦0) + C(2¦1) + C(2¦2) = 4. Nope.
Also note that ∑C(n¦k), [with k=0,1,2,...,n] equals 2^n.

Therefore we have: 16 = 2^n from which n=4
Hence, it is also sufficient.

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