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Your math club has 20 members. In how many ways can it select a president and a vice-president if no member can hold more than one office?

 Nov 19, 2020
 #1
avatar+2225 
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The formula is for combinations: \(\frac{n!}{k!(n-k)!}\)

Plugging in, we get

20!/(2! times 18!) which will get you 190.

 

So, the answer is 190 ways.

 Nov 19, 2020
 #2
avatar+116125 
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We can choose  any 1 of the 20 members for  president

 

Once any of these is selected, we can choose any 1of the remaining 19 members for vice-president

 

So   .....by the fundamental counting principle  =  20 * 19   =   380  ways

 

 

cool cool cool

 Nov 19, 2020
 #3
avatar+2225 
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Chris, I don't get how my answer is incorrect. Will you mind explaining why it is wrong?

CalTheGreat  Nov 19, 2020
 #4
avatar+116125 
+1

OK, Cal....you ask  a good question....!!!!

 

Let's  look  at something  more simple

 

Suppose  we only had 4 members    A, B C D

 

Let   A/B A/C A/D  B/C B/D C/D      represent     president / vice-president

 

Note   that  this is   C(4,2)  =       4!

                                                 _______     =       6

                                                 (4 - 2)! 2!

 

But....what we have  as actually a permutation     because   A/B    isn't the same as  B /A

 

So   we actually have   P(4,2)  =   12

 

So  your answer  is a combination  C(20,2) = 190

 

But we should have a permutation =   P(20, 2)  = 380......  this is the same result as 20 * 19

 

 

 

 

 

cool cool cool

CPhill  Nov 19, 2020
 #5
avatar+2225 
0

I get it, thanks a lot Chris!! laugh

 Nov 19, 2020

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