Question

1. oliver (cat) and dodger (dog) belong to a club of cats and dogs. there are five dogs and six cats in the club. for each case below: how many ways can a committee of four be selected to work on finding food next saturday?

a) there must be two dogs and two cats. (2 marks)
b) there must be at least two cats. (3 marks)
c) oliver and dodger must be on the committee. (2 marks)
d) there must be more dogs than cats. (3 marks)

Explanation:

Step1: Calculate combinations for part a

We use the combination formula \(C(n,r)=\frac{n!}{r!(n - r)!}\). For dogs, \(n = 5\), \(r=2\), so \(C(5,2)=\frac{5!}{2!(5 - 2)!}=\frac{5\times4}{2\times1}=10\). For cats, \(n = 6\), \(r = 2\), so \(C(6,2)=\frac{6!}{2!(6 - 2)!}=\frac{6\times5}{2\times1}=15\). Then multiply the two results: \(10\times15 = 150\).

Step2: Calculate combinations for part b

Case 1: 2 cats and 2 dogs. \(C(6,2)\times C(5,2)=15\times10 = 150\). Case 2: 3 cats and 1 dog. \(C(6,3)=\frac{6!}{3!(6 - 3)!}=\frac{6\times5\times4}{3\times2\times1}=20\), \(C(5,1)=\frac{5!}{1!(5 - 1)!}=5\), product is \(20\times5=100\). Case 3: 4 cats and 0 dogs. \(C(6,4)=\frac{6!}{4!(6 - 4)!}=\frac{6\times5}{2\times1}=15\), \(C(5,0)=1\), product is \(15\times1 = 15\). Sum the cases: \(150+100 + 15=265\).

Step3: Calculate combinations for part c

Since Oliver and Dodger are already on the committee, we need to choose \(4-2 = 2\) more members from \(5 + 6-2=9\) remaining. \(C(9,2)=\frac{9!}{2!(9 - 2)!}=\frac{9\times8}{2\times1}=36\).

Step4: Calculate combinations for part d

Case 1: 3 dogs and 1 cat. \(C(5,3)=\frac{5!}{3!(5 - 3)!}=\frac{5\times4}{2\times1}=10\), \(C(6,1)=6\), product is \(10\times6 = 60\). Case 2: 4 dogs and 0 cats. \(C(5,4)=\frac{5!}{4!(5 - 4)!}=5\), \(C(6,0)=1\), product is \(5\times1=5\). Sum the cases: \(60 + 5=65\).

Answer:

a) \(150\)
b) \(265\)
c) \(36\)
d) \(65\)