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I have a set like this: A = {1, 2, 3, 5}. From this set, I can create many set like this:

  • Only one member in a new set: {1}, {2}, {3}, {5}. And the times a member occurs is equal to 1. ( Yes, 1, 2, 3, and 5 just occurs 1 times).
  • 2 members in a new set: {1, 2}, {1, 3}, {1, 5}, {2, 3}, {2, 5}, {3, 5}. The number of set is equal to 4C2 = 4!/(2!(4-2)!) = 6. And the times a member occurs is equal to 3. (Yes, each of member occurs 3 times, you can count it).
  • 3 members in a new set: {1, 2, 3}, {1, 2, 5}, {1, 3, 5}, {2, 3, 5}. The number of set is equal to 4C3 = 4!/(3!(4-3)!) = 4. Similar, the time a member occurs is equal to 3 times.
  • 4 members in a new set: {1, 2, 3, 5}. The number of set equal to 4C4 = 4!/(4!(4-4)!) = 1. Similar, the time a member occurs is equal to 1 times.

But in this case, my set (A) is small. When it large(with n members, n is natural number), how can I count how many times a member occurs in a each of many new set.

Sorry for my bad english. Hope you can understand and give me any ideas or any solutions. Thanks so much.

Andrés E. Caicedo
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yanndinh
  • 13

1 Answers1

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Suppose that $|A| = n$. Let $a \in A$. We wish to determine how many subsets of $A$ with cardinality $k$ contain $A$. To form a subset $B$ of $A$ such that $a \in B$ and such that $|B| = k$, we must select $k - 1$ of the other $n - 1$ elements of set $A$, which can be done in $$\binom{n - 1}{k - 1}$$ ways.

Let's verify this for your example. If $|A| = 4$, then the number of ways an element can occur in a subset of size

  • $1$ is $\binom{4 - 1}{1 - 1} = \binom{3}{0} = 1$
  • $2$ is $\binom{4 - 1}{2 - 1} = \binom{3}{1} = 3$
  • $3$ is $\binom{4 - 1}{3 - 1} = \binom{3}{2} = 3$
  • $4$ is $\binom{4 - 1}{4 - 1} = \binom{3}{3} = 1$

as you found.

N. F. Taussig
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