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My name is Ann [436]
3 years ago
9

Let abe a finite set and let a∈a. prove that the number of subsets of a that contain a1 equals the number of subsets of a that d

o not contain a1.
Mathematics
1 answer:
vladimir1956 [14]3 years ago
5 0
Here's a combinatorial proof. Suppose A has n elements.

For a subset to contain a, it must consist of at least one element. So if any given subset has k elements, where 1\le k\le n, then a is not one of the other k-1 elements. This means the number of subsets containing a is

\displaystyle\sum_{k=1}^n\binom11\binom{n-1}{k-1}

Put another way, we are choosing elements from A to form a subset of k elements. We want a to be in each subset, so we have n-1 other elements of A from which to choose. Then we sum over all the possible sizes of the desired subset.

On the other hand, if we want to build subsets not containing a, then we have n-1 total elements to choose from, and we can make subsets of size ranging from 0 to n-1, so the number of subsets not containing a is

\displaystyle\sum_{k=0}^{n-1}\binom10\binom{n-1}k

We have \dbinom10=\dbinom11=1, and in the second sum we can shift the index up by 1 to get

\displaystyle\sum_{k=1}^{n-1+1}\binom10\binom{n-1}{k-1}

which is the same as the first count.
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