+0  
 
0
35
1
avatar+418 

My school's Physics Club has 22 members. It needs to select 3 officers: chairman, vice-chairman, and sergeant-at-arms. Each person can hold at most one office. Two of the members, Penelope and Quentin, will only be officers if the other one is also an officer. (In other words, either both Penelope and Quentin are officers, or neither are.) In how many ways can the club choose its officers?

Logic  Oct 19, 2018
 #1
avatar+2729 
+1

\(\text{Let's count the combination that have neither Penelope or Quentin first}\\ \text{there will be }\dbinom{20}{3}3! =6840 \text{ ways to assign the officers w/o P or Q}\\ \text{now if we have P we have Q so that just leaves }\dbinom{20}{1}=20 \text{ combinations}\\ \text{or }20\cdot 3!=120 \text{ different ways off assigning the officers using P and Q}\\ \text{this gets us a total of }6960 \text{ ways of assigning officers}\)

Rom  Oct 19, 2018

34 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.