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How many non-empty subsets of $\{ 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \}$ consist entirely of prime numbers? (We form a subset of the group of numbers by choosing some number of them, without regard to order. So, $\{1,2,3\}$ is the same as $\{3,1,2\}$.)

Guest Sep 21, 2017
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How many non-empty subsets of \(\{ 1 , 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 \}\) consist entirely of prime numbers?

(We form a subset of the group of numbers by choosing some number of them, without regard to order.

So, \(\{1,2,3\}\) is the same as \(\{3,1,2\}\).)

 

The set of prime numbers \( \{ 2, 3, 5, 7, 11 \}\) consist of 5 elements.

 

The subsets with one element \( = \binom{5}{1} = 5 \) subsets.
The subsets with two elements \(= \binom{5}{2} = 10\) subsets.
The subsets with three elements \(= \binom{5}{3} = 10\) subsets.
The subsets with four elements \(= \binom{5}{4} = 5\) subsets.
The subsets with five elements \(= \binom{5}{5} = 1\) subset.

 

The sum is 5+10+10+5+1 = 31 subsets

 

laugh

heureka  Sep 21, 2017

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