A catering service offers 3 appetizers, 7 main courses, and 4 desserts. A customer is to select 2 appetizers, 3 main courses,and 3 desserts for a banquet. In how many ways can this be done?

The combination formula C(n,r) shows us the number of ways of picking r unordered outcomes from n possibilities. For n ≥ r ≥ 0, C(n,r) is given by the following formula:

[tex]C(n,r)=\frac{n!}{r!(n\text{ -r})!}[/tex]

Explanation:

We can use combinations and fundamental counting principles to answer this question.

Let us denote by "a", the number of selection of appetizers. This number can be calculated as follows:

[tex]C(3,2)=\frac{3!}{2!(3\text{ -2})!}=3[/tex]

Now, let us denote by "b", the number of selection of main courses. This number can be calculated as follows:

[tex]C(7,3)=\frac{7!}{3!(7\text{ -3})!}=35[/tex]

Finally, let us denote by "c", the number of selection of desserts. This number can be calculated as follows:

[tex]C(4,3)=\frac{4!}{3!(4\text{ -3})!}=4[/tex]

Now, applying the multiplication principle we get the desired number:

[tex]C(3,2)\cdot C(7,3)\cdot C(4,3)=3\cdot35\cdot4\text{ = 420}[/tex]

We can conclude that the correct answer is: