How many ways can we choose four subsets of $3$ elements from {1, 2, .., 6} so that each pair of subsets share exactly one element?
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Each pair of subsets share exactly one element
combinatorics
Best Answer
The first subset can be any three elements, so six-choose-three equals $20$ possibilities. Suppose it's $\{\,1,2,3\,\}$. The second subset must have one of these three elements, and two of the three elements not in the first subset, so $3\times3=9$ ways to choose the second subset. Let's say it's $\{\,1,4,5\,\}$. The third subset must have $2$ or $3$, and $4$ or $5$, and $6$, so, four ways to choose the third subset. Say it's $\{\,2,4,6\,\}$. Then the fourth subset must be $\{\,3,5,6\,\}$. So we have $20\times9\times4=720$ ways. BUT it doesn't matter which of the four subsets is first, which is second, and so on, so we have to divide by $24$.