A’ will contain how many elements from the original set A
Answers
Answer:
The first case is quite simple. You have the set A={1,2,3,4} and the set B={2,3,4}. Now, every subset of size m of A which contains 1 can be formed by joining a generic subset of B of size m−1 with the set {1}.
In other word, let n=4 and m=2 (as you propose), then (n−1)!(m−1)!(n−1−m+1)!=3!1!2!=3 is the number of the subset of size 1 of B and then this is the number of subset of A of size 2 which contains 1.
I guess the second case can be solved with similar arguments. Just to give an idea...
Let A be a set (without repeated elements). For example A={1,2,4}.
Suppose that each element of A has a multiplicity, say μ(x) where x is an element of A . In the example, μ(1)=3, μ(2)=μ(4)=1. Suppose also that you remove an element x from the set A and you obtain the set B=A∖{x}. Let nA=μ(x) and nB=∑y≠xμ(y). Clearly nA+nB=n.
When you have to form a group of m element extracted from A (with repeated elements) then you can say that:
I extract m elements from collection B (if m≤nB)
I extract m−1 elements from collection B and 1 from collection {x} (if m−1≤nB and 1≤nA)
I extract m−2 elements from collection B and 2 from collection {x} (if m−2≤nB and 2≤nA)
and so on...
and so on
Each case can be easily computed using binomial coefficients so the solution is straight forward!
NOTE I use collection to distinguish from set. A set has unique elements, a collection can have repeated elements. In my answer I use a set or a collection depending from the context.
A as set: A={1,2,4}
A as collection: A={1,2,1,1,4}
B as set: B={2,4}
B as collection: B={2,4}
{x} as set: {x}={1}
{x} as collection: {x}={1,1,1}