Statement 1 seems incomplete. How many people on each list ? If nit equal, what sort of distribution. With the current information, you could have infinite no of atleast 2 disciples distributed on 200-500 lists.

Statement 2: nC2=28 => n!/(n-2)!2! = 28 => n(n-1)=56 => n^2-n-56=0 => n^2-8n+7n-56=0
=> n(n-8)+7(n-8)=0 =>(n+7)(n-8)=0

n=-7 not possible. So n=8.

Hence B

sana.noor wrote:How many disciples does Tiffany have?

(1) there are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples.

(2) there are 28 ways that she could decide which 2 disciples she will recommend promoting.

OA is D

Statement 2: There are 28 ways that she could decide which 2 disciples she will recommend promoting.
Number of ways to choose 2 people from n choices = (n)(n-1) / (2*1).
Since there are 28 ways, we get:
(n)(n-1) / (2*1) = 28.
(n)(n-1) = 56.
Since n must be a positive integer, n=8, implying that there are 8 disciples.
SUFFICIENT.

Statement 1: There are between 200 and 500 lists she could make consisting of the names of at least 2 of her disciples.
Since the two statements cannot contradict each other, it must be possible that 8 disciples will satisfy the constraint in statement 1.
The question at hand is whether there can be MORE THAN 8 or FEWER THAN 8 disciples.

Case 1: 9 disciples
For each disciple, there are two options: to be included on a list or NOT to be included.
Since there are 2 options for each of the 9 disciples, we get:
2*2*2*2*2*2*2*2*2 = 512.
Of these 512 lists, the following cases violate the constraint that at least 2 disciples must be included:
Number of ways to choose NO disciples = 1.
Number of ways to choose exactly ONE disciple = 9.
Thus:
Total number of possible lists = 512-1-9 = 502.
Since 502 is not between 200 and 500, there cannot be 9 disciples.

Case 2: 7 disciples
Applying the reasoning used in Case 1, the total number of possible lists ≈ 2*2*2*2*2*2*2 = 128.
Since the total number of possible lists will not be between 200 and 500, there cannot be 7 disciples.

Thus, just like statement 2, statement 1 requires that the number of disciples = 8.
SUFFICIENT.

The correct answer is D.

We're getting into deep waters here, but I wanted to explain the trick used for the first statement and make it a little easier to work with.

If I have a group of x disciples to choose from, the sum of the ways to choose all the subgroups from 0 of the disciples to all x of the disciples is 2^x.

(This formula is intuitive enough: for every member of the group, there are two choices - SELECTED or NONSELECTED - and since we have x disciples to consider, our formula is 2^x. A proper proof is here.)

Since we want to know the number of ways of choosing at least 2 of the members, we want to subtract the number of ways of choosing 0 disciples and the number of ways of choosing 1 disciple.

x choose 0 = 1 (there's only one way to choose 0 disciples)
x choose 1 = x (also pretty straightforward)

So the number of ways of choosing at least 2 of x disciples is 2^x - x - 1.

Statement 1 gives us the inequality

200 < 2^x - x - 1 < 500
201 < 2^x - x < 501

The only power of 2 in this range at all is 2^8, so x = 8, and S1 is SUFFICIENT.