At a local restaurant a new dinner menu is available that offers 4 courses 1 appetizer 1 salad 1 main dish and 1 dessert all for one price. there are 6 appetizers 2 salads 3 main dishes and 4 desserts. how many dinner outcomes are there possible

Number of Dinner Menu in the Restaurant = 4 =[1 appetizer+ 1 salad+ 1 main dish + 1 dessert]

Number of Different Appetizer = 6

Number of Different Salad =2

Number of Different Main Dishes = 3

Number of Different Desserts =4

Number of Different Dinner Outcomes

     =[Choosing 1 appetizer among 6 appetizer] × [Choosing 1 Salad item among 2 salad Items] ×[Choosing 1 main Dish among 3 main Dishes] × [Choosing 1 dessert among 4 desserts]

=Order is not Important,so we will use the concept of combinatorics

So, Number of Possible Dinner Outcomes = 144

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )