Question

A veggie wrap at Dylan's Diner is composed of 3 different vegetables and 3 different condiments wrapped up in a tortilla. If there are 9 vegetables, 5 condiments, and 5 types of tortilla available, how many different veggie wraps can be made

Answer 1

Answer:

$5C1 (9C3) (5C3)$

And replacing we got:

$\frac{5!}{1! 4!} (\frac{9!}{3! (9-3)!}) (\frac{5!}{3! (5-3)!}) = 5*84*10=4200$

So then we have a total of 4200 possibilities in order to have a veggle wrap

Step-by-step explanation:

For this case we can use the combinatory formula given by:

$nCx= \frac{n!}{x! (n-x)!}$

And for this case we have a total of 9 vegatables and 5 condiments and 5 types of tortilla. Then the total number of possibilities are:

$5C1 (9C3) (5C3)$

And replacing we got:

$\frac{5!}{1! 4!} (\frac{9!}{3! (9-3)!}) (\frac{5!}{3! (5-3)!}) = 5*84*10=4200$

So then we have a total of 4200 possibilities in order to have a veggle wrap