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Ten people, including Fred, are in the Jazz Club. They decide to form a 3-person steering committee. How many possible committees can be formed that do not include Fred? Do you notice a relationship between your answers of the three parts of this problem?

Guest Mar 20, 2015

Best Answer 

 #1
avatar+80935 
+11

Using my two previous answers.....

Total committees - Total committtees that include Fred = Total committees that don't include Fred

So we have

120 - 36  =  84 committees that don't include Fred

Note this is the same as C(9,3) = choosing  any 3 of 9 people that don't include Fred

 

  

CPhill  Mar 20, 2015
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2+0 Answers

 #1
avatar+80935 
+11
Best Answer

Using my two previous answers.....

Total committees - Total committtees that include Fred = Total committees that don't include Fred

So we have

120 - 36  =  84 committees that don't include Fred

Note this is the same as C(9,3) = choosing  any 3 of 9 people that don't include Fred

 

  

CPhill  Mar 20, 2015
 #2
avatar+91435 
+5

total number of commities - no restrictions         $${\left({\frac{{\mathtt{10}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{10}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{120}}$$

 

total number that Fred is in         $${\left({\frac{{\mathtt{9}}{!}}{{\mathtt{2}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{2}}){!}}}\right)} = {\mathtt{36}}$$

 

 

total number that Fred is not ine       $${\left({\frac{{\mathtt{9}}{!}}{{\mathtt{3}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{3}}){!}}}\right)} = {\mathtt{84}}$$

 

$${\mathtt{84}}{\mathtt{\,\small\textbf+\,}}{\mathtt{36}} = {\mathtt{120}}$$

These 2 are complementary events,

either one happens OR the other happens. 

Melody  Mar 21, 2015

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