if the members are different then total subsets is equal to 15C1 + 15C2 +....15C15
this is the usual notation of combinatorics.
and nCr is equal to fact(n)/{ fact(r) multiply fact(n-r)}
and Fact(m) = m*(m-1)*(m-2)......3*2*1
A set with 50 elements has 250 subsets.
In a set, as it is usually defined, elements can't be repeated. "Mathematics" has 8 distinct letters, so your set would have 8 letters. The number of possible subsets (this includes the empty set, and the set itself) is two to the power 8.
64. You can use Pascal's triangle to figure out how many subsets have no elements, one element, two elements and so on. For this particular one, you will have 6 subsets with one element, 15 with two, 20 with three, 15 with four, 6 with five and only one each of all six and none at all.
That is the same as asking 15C10 which is 3003. It is 15! divided by (15-10)!(10!) This is 15x14x13x12x11 divided by 5x4x3x2x1 or divided by 120 or 360360=3003
Well, honey, I hope you're ready for this math lesson. A set with 6 elements can have 2^6, which is 64 subsets. That's right, 64 ways to slice and dice those elements. So, grab a calculator and start counting, darling!
The one where the highest and lowest members differ by 15.
15
15
15
The United Nations Security Council has 15 members.
15
15