Respuesta :
Answer:
24 possible dishes
Step-by-step explanation:
Given
Total
[tex]Meat = 4[/tex]
[tex]Cheese = 3[/tex]
[tex]Bread = 2[/tex]
Selection
[tex]Meat = 1[/tex]
[tex]Cheese =2[/tex]
[tex]Bread = 1[/tex]
Required
Determine the number of possible combo
Rita wants to select 1 of 4 meat types,
So, the number of selections is: [tex]^4C_1[/tex]
Rita wants to select 2 of 3 cheese types,
So, the number of selections is: [tex]^3C_2[/tex]
Rita wants to select 1 of 2 cheese types,
So, the number of selections is: [tex]^2C_1[/tex]
The total number of selections is:
[tex]Total = ^4C_1 * ^3C_2 * ^2C_1[/tex]
Using
[tex]^nC_r = \frac{n!}{(n-r)!r!}[/tex]
We have:
[tex]Total = \frac{4!}{(4-1)!1!} * \frac{3!}{(3-2)!2!} * \frac{2!}{(2-1)!1!}[/tex]
[tex]Total = \frac{4!}{3!1!} * \frac{3!}{1!2!} * \frac{2!}{(1!1!}[/tex]
[tex]Total = \frac{4*3!}{3!*1} * \frac{3*2!}{1*2!} * \frac{2*1}{1*1}[/tex]
[tex]Total = 4* 3 * 2[/tex]
[tex]Total = 24[/tex]
Hence, there are 24 possible dishes