+0  
 
0
521
1
avatar

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
 #1
avatar+20153 
+1

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

8 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.