**Exercise #1: Restaurant Menu Appetizers Nachos …………………… 120.00 Salad ……………………… 125.00 Main Courses Hamburger ………………. 125.00 Cheeseburger …………… 130.00 Fish Filet …………………. 155.00 Beverages Tea ……………………….. 120.00 Milk ………………………. .120.00 Cola ………………………. 110.00 Root Beer ………………... 110.00 a. Consider Illustration above. Find the number of dinner that can be served if we have one appetizer, one main course and one beverage, we can again use the multiplication principle. b. How many strings of length four can be formed using the letters A, B, C, D, E? c. How many strings of length four can be formed using the letters A, B, C, D, E if the string begins with letter C?**

The **Answer to the Question**

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**Here's the Solution to this Question**

a. There are two Appetizers, three Main Courses, and four Beverages.

Use the multiplication rule

$2\cdot3\cdot4=24$

24 kinds of dinner can be served.

b.

If repetitions are allowed

$5^4=625$

625 strings of length four can be formed using the letters A, B, C, D, E if repetitions are allowed.

If repetitions are not allowed

$P(5, 4)=\dfrac{5!}{(5-4)!}=120$

120 strings of length four can be formed using the letters A, B, C, D, E if repetitions are not allowed.

c.

If repetitions are allowed

$5^3=125$

125 strings of length four can be formed using the letters A, B, C, D, E if the string begins with letter C and repetitions are allowed.

If repetitions are not allowed

$P(4, 3)=\dfrac{4!}{(4-3)!}=24$

24 strings of length four can be formed using the letters A, B, C, D, E if the string begins with letter C and repetitions are not allowed.