PLEASE NO HAND WRITTEN THANK YOU!!!! 2. At the Phoenix annual meeting of the pet
ID: 3147214 • Letter: P
Question
PLEASE NO HAND WRITTEN THANK YOU!!!!
2. At the Phoenix annual meeting of the pet owner's club, a show of hands determined that 41 people in attendance owned one or more dogs but no cat; 38 owned one or more cat but no dogs; and 12 owned both types of pei A dog/cat owner committee of 6 people is to be formed. It is thought that the diversity of the dog/cat owning community is best represented if 2 people on that committee only own a dog or dogs; 2 people only own a cat or cats; and 2 people own both dog(s) and cat(s). In how many ways can this committee be selected? Explain your reasoning and show work. 3. How does the solution to the previous problem change if the two people selected out of each of the three distinct groups do not have equal power on the committee, but rather, the first selected is the "speaker for that group, and the second selected is just an advisor? Explain your answerExplanation / Answer
Please ask if any doubts
Given data
No of people only with dog/dogs=41
No of people with cat=38
No of people with both type of animal=12
So we know that we have to select 2 person from each group
So no of ways to select 2people from 41=41!/2!.39!
=41*40/2
=820
Now same for cat=38!/2!.36!
=38*37/2
=703
Now from both type it will be
12!/2!.10!
=66
So toal no of ways=66*703*820
2.it will be like this
No of ways to select 2 peolpe who only owns dog/dogs=2*41!/2!.39!
=41*40
=1640
No of ways 2pesron with cat/cats can be select=38!/2!*36!
=38*37
=1406
No if ways 2 people can be selected who has bith pets=12!*2/10!*2!
=132
So total no of ways=1640*1406*132
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