Solution requested to probability question

[ewout]

Could somebody provide the answer to the following two questions?

1) Why is the following question a permutation instead of a combination? 'In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?'

2) Could you please explain how to solve the following question with a (factorial) formula? 'A menu offers 2 entrees, 3 main courses, and 3 desserts. How many different combinations of dinner can be made? (a dinner must contain an entree, a main course and a dessert)'

 

[galactus]

1) Why is the following question a permutation instead of a combination? 'In how many ways can 5 boys and 4 girls be arranged in a line so that there will be a boy at the beginning and at the end?'

Because you have to choose two boys from the 5, and which boy goes on the end matters. i.e. First, choose 2 boys to go on the ends.

\(\displaystyle C(5,2)=10\). But, they can be switched around and placed in 2 ways. Hence, \(\displaystyle 2C(5,2)=\frac{2\cdot 5!}{2!(5-2)!}=20\).

Using a permutation, \(\displaystyle P(5,2)=\frac{5!}{(5-2)!}=20\)

Now, arrange the 4 girls and 3 remaining boys in the line between the two boys.

2) Could you please explain how to solve the following question with a (factorial) formula? 'A menu offers 2 entrees, 3 main courses, and 3 desserts. How many different combinations of dinner can be made? (a dinner must contain an entree, a main course and a dessert)'

No need for factorials. Just multiply them together. \(\displaystyle 2\cdot 3\cdot 3\)

Or, choose 1 entree from 2, 1 main course from 3, and 1 dessert from the 3

\(\displaystyle \text{entree}=\frac{2!}{1!(2-1)!}=2\), and so on.