Let S be a subset of {1, 2, 3,... 100}, containing 50 elements. How many such sets have the property that every two numbers in S have a common divisor that is greater than 1?

I really need help. Give me a good explanation with he answer. Thanks!!! :)

Let S2 be the subset containing all the multiples of 2. There are 50 of them, and they all have 2 as a divisor. So, S2 is one of our subsets.

There are fewer than 50 multiples of any other number in a subset of {1 ... 100}.

Looks like S2 is our only candidate.

To find the number of such sets, let's start by thinking about the condition mentioned - "every two numbers in the set have a common divisor greater than 1." This implies that every pair of numbers in the set must have at least one prime number in common.

Let's consider an arbitrary pair of numbers from the set, say a and b. For the condition to be satisfied, both a and b must have a common prime divisor. Now, let's break the problem into two cases:

Case 1: Both a and b are odd numbers
In this case, the only common divisor they can have is 1, as they cannot share any common prime factors. Hence, for the condition to be met, we cannot have two odd numbers in the set.

Case 2: One or both of a and b are even numbers
In this case, the common divisor can be 2 since all even numbers have 2 as a common factor. Therefore, to satisfy the condition, we need to include at least one even number in the set.

Now, let's count the number of sets for each case:

Case 1: There are 50 odd numbers between 1 and 100. The number of sets without any odd numbers is the number of ways to choose 50 even numbers from the remaining 50 even numbers (from 2 to 100) out of which 50 can be chosen. This can be denoted as C(50, 50), which is equal to 1.

Case 2: There are 50 even numbers between 1 and 100. We can choose any of the 50 even numbers to include in the set. Therefore, the number of sets for this case is 50.

Finally, to find the total number of sets, we sum the number of sets from each case:

Total number of sets = number of sets in Case 1 + number of sets in Case 2
= 1 + 50
= 51

So, there are 51 such sets that satisfy the given condition.