Official ExplanationFind the number of choices for appetizers with the
“choose” function, written as and defined for positive integers by the formula

Recall that n! means the product of all positive integers from 1 through n, inclusive. From six appetizers, we choose two, so there are 6C2 combinations, which equals

after canceling duplicate numbers in the numerator and denominator. You can also compute this by
“brute force.” Label the appetizers A, B, C, D, E, and F, and then list all combinations of two appetizers as follows: AB, AC, AD, AE, AF, BC, BD, BE, BF, CD, CE, CF, DE, DF, EF. There are 5 + 4 + 3 + 2 + 1 = 15 combinations in all. Multiply the number of choices for appetizer by the four choices for entrée and the five choices for dessert to obtain 15 × 4 × 5 = 300 total menu combinations.
Answer: 300Attachment:
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