Exam (elaborations)

Assignment/Test (elaborations) MDM4U

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Course
MDM4U

Institution
12th Grade

This assignment is designed to deepen your understanding of permutations and counting, essential concepts in combinatorics and data management. Through a series of engaging problems and practical applications, you'll explore the fundamental principles of arranging, ordering, and selecting objects w...

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Uploaded on August 9, 2023
File latest updated on August 9, 2023
Number of pages 16
Written in 2023/2024
Type Exam (elaborations)
Contains Questions & answers

calculus
maths
12th grade
university preparation
mdm4u permutation and counting

Institution Secondary school
Education 12th Grade
Course MDM4U
School year 1

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0.1. Assignment: Organized counting

a. Permutation- It is an arrangement of objects where order matters. For
example, arranging the letters 𝐴𝐵𝐶 in different ways such as 𝐴𝐵𝐶, 𝐴𝐶𝐵, 𝐵𝐴𝐶
etc.
b. Combination- It is an arrangement of objects where order does not
matter. For example, choosing members of a committee.
c. Factorial- It is the product of a number and all the positive integers
below it. For example, 7 's factorial is 7 ∗ 6 ∗ 5 ∗ 4 ∗ 3 ∗ 2 ∗ 1

d. Mutually exclusive- It is a term used to describe the kind of events
that cannot occur at the same time. Only one can take place at a time. For
example- flipping a coin. It can either be heads or tails. It cannot be both at
the same time.
2a. Create a tree diagram that depicts all possible options for Chris'
meal.

b. How many different meals are possible?

18 meals are possible.
Since there are 3 appetizers, 2 main courses and 3 dessert options, according
to fundamental counting principle.

3 ∗ 2 ∗ 3 = 18 meals

,c. How many different meals are possible if Chris cannot serve Eggplant
and Fritter together? Explain your answer.

If eggplants and fritters cannot be served together, the following 3 meals
would not exist.
Artichoke-Eggplant-Fritter Bruschetta-Eggplant-
Fritter Caviar-Eggplant-Fritter.

Since the total number of meals is 18 and now that we cannot serve 3 of
them, we would have 18-3=15 possible meals.
d. How many different meals are possible if Chris must serve at least one
of Artichoke, Eggplant, and Hot chocolate?
If Chris must serve at least one of Artichoke, eggplant, and hot chocolate the
number of possible melas would decrease by 4 (BDF, BDG, CDF, CDG).

, As you can see in the image, the total number of meals that would have
Artichoke as an appetizer are 6 , so we don't need to eliminate any meal
from there.

Moving onto Bruschetta, if the meal has eggplant as main course, we can
keep all three meals but if the meal has drumsticks as the main course, we
can only keep one meal that has hot chocolates as the dessert, adding up to 4
possible meals.
Similarly, the total number of meals with Caviar as the appetizer would have
to eliminate two of the meals. Leaving us with 4 meals total.

After adding up all the numbers we will have 6+4+4=14 OR we can
eliminate the bad meals from the total number of possible meals, 18-4=14.

So, the answer would be 14 meals.
3. A teacher is creating a test from a question bank. If question 1 can be
chosen from 8 options, question 2 chosen from 3 options, question 3
from 2 options, and question 4 from 21 options, use the multiplicative
counting principle to determine the number of different tests the
teacher could make.

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