Question

A restaurant lunch special allows the customer to choose three vegetables from the following group. peppers carrots radishes broccoli fiddleheads cauliflower okra corn. How many outcomes are possible if the customer chooses three different​ vegetables?

Answer 1

Answer:

There are 56 outcomes are possible if the customer chooses three different​ vegetables.

Step-by-step explanation:

Given that the vegetables are peppers ,carrots, radishes ,broccoli, fiddle heads, cauliflower, okra and corn.

There are 8 different vegetables in the given group.

A restaurant lunch special allows the customer to choose 3 vegetables from the given group.

To find how many outcomes are possible if the customer chooses three different​ vegetables :

From the given data the customer has to choose 3 different vegetables from the group.

So, total number of possibilities of selecting 3 vegetables can be solved by Combinations ($^nC_r$)

Here n=8 and r=3

The formula is $^nC_r=\frac{n!}{(n-r)!r!}$

Substitute the values in the formula we get

$^8C_3=\frac{8!}{(8-3)!3!}$

We know $n!=n(n-1)(n-2)...3.2.1$

$^8C_3=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{(5)!3!}$

$^8C_3=\frac{8\times 7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1(3\times 2\times 1)}$

$=\frac{56}{1}$

$=56$

∴ $^8C_3=56$

∴ there are 56 outcomes are possible if the customer chooses three different​ vegetables.