You are watching: How many subsets does the set {1, 2, 3} have?

I nothing think I recognize the solution.

According to the systems if i take out $n-1$ native $1,2,3,ldots , n-1, n$, then ns would have $S_n-2$ come the left that $n-1$, which i get, yet then I have ‘$n$’ which i don’t know what to do with. Because that every subset connected with $S_n-2$, there’s currently an element ‘$n$’ easily accessible to either incorporate or not include in the subset, which boosts the size of $S_n-2$. If pretend ‘$n$’ isn’t there it would give us $S_n-2$ exactly.

The very same goes for $S_n-2$ and also $S_n-3$. The $S_n$ would certainly then it is in a reduced bound.

What am I no getting around this question?

combinatorics recurrence-relations

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edited Dec 13 "19 at 6:28

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inquiry Dec 12 "19 at 23:58

noname123noname123

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## 1 answer 1

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Let $A subseteq 1, 2, cdots, n $ satisfies the condition.We deserve to divide to 3 cases:$n ot in A$: Then, there room $S_n-1$ possibilities because that $A$$n in A$:$n-1 ot in A$ : Then, there room $S_n-2$ possibilities that $A$ because $A - $ is three-consecutive-integer-free and also belongs come $1, 2, cdots, n-2$.$n-1 in A$: Then, $n-2 ot in A$. Currently we have actually $A - -1, n subset 1, 2, cdots, n-3$, therefore there room $S_n-3$ possibilities because that $A$.

Therefore, $S_n = S_n-1 + S_n-2 + S_n-3$.

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answer Dec 13 "19 in ~ 4:33

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