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16 June, 19:06

For dinner, a restaurant allows you to choose either Menu Option A: five appetizers and three main dishes or Menu Option B: three appetizers and four main dishes. There are six kinds of appetizer on the menu and five kinds of main dish.

How many ways are there to select your menu, if ...

a. You may not select the same kind of appetizer or main dish more than once.

b. You may select the same kind of appetizer and/or main dish more than once.

c. You may select the same kind of appetizer or main dish more than once, but not for all your choices, For example in Menu Option A, it would be OK to select four portions of 'oysters' and one portion of 'pot stickers', but not to select all five portions of 'oysters'.)

In each case show which formula or method you used to derive the result.

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Answers (1)
  1. 16 June, 20:55
    0
    The formula used in this question is called the probability of combinations or combination formula.

    Explanation:

    Solution

    Given that:

    Formula applied is stated as follows:

    nCr = no of ways to choose r objects from n objects

    = n! / (r! * (n-r) !)

    The Data given:

    Menu A : 5 appetizers and 3 main dishes

    Menu B : 3 appetizers and 4 main dishes

    Total appetizers - 6

    Total main dishes - 5

    Now,

    Part A:

    Total ways = No of ways to select menu A + no of ways to select menu B

    = (no of ways to select appetizers in A) * (no of ways to select main dish in A) + (no of ways to select appetizers in B) * (no of ways to select main dish in B)

    = 6C5*5C3 + 6C3*5C4

    = 6*10 + 20*5

    = 160

    Part B:

    Since, we can select the same number of appetizers/main dish again so the number of ways to select appetizers/main dishes will be = (total appetizers/main dishes) ^ (no of appetizers/main dishes to be selected)

    Total ways = No of ways to select menu A + no of ways to select menu B

    = (no of ways to select appetizers in A) * (no of ways to select main dish in A) + (no of ways to select appetizers in B) * (no of ways to select main dish in B)

    = (6^5) * (5^3) + (6^3) * (5^4)

    = 7776*125 + 216*625

    = 1107000

    Part C:

    No of ways to select same appetizers and main dish for all the options

    = No of ways to select menu A + no of ways to select menu B

    = (no of ways to select appetizers in A) * (no of ways to select main dish in A) + (no of ways to select appetizers in B) * (no of ways to select main dish in B)

    = (6*5) + (6*5)

    = 60

    Total ways = Part B - (same appetizers and main dish selected)

    = 1107000 - 60

    = 1106940
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