A Harry Potter - Themed Quiz on Combinatorics | Sci / Tech

Quiz Answer Key and Fun Facts

1. In her Arithmancy class, Hermione was asked to form 5 - digit even numbers from the numbers 1, 4, 5, 8, and 0. How many can she make if repetition of numbers is not allowed?

Answer: 60

Case 1: 0 is ones digit --> (place value of digit : # of choices)
(tth : 4) (th : 3) (h : 2) (t : 1) (o : 1)-> 4 x 3 x 2 x 1 x 1 = 24
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Case 2: 4 or 8 is ones digit :
(tth : 3) (th : 3) (h : 2) (t : 1) (o : 2) -> 3 x 3 x 2 x 1 x 2 = 36
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Case 1 + Case2 = 24 + 36 = 60 ways

Taking Cases was used by Hermione in this problem because of the conflict with 0. Zero cannot be included in the choice for the ten thousands digit because the resulting number will only have 4 digits.

2. Suppose Professor Mc Gonagall was making a list of possible passwords for the Gryffindor Tower. If she were to make passwords of 7 distinct letters from the letters of the English alphabet, how many of these passwords will have "A" in the middle?

Answer: 127,512,000

Prof Mc Gonagall did the following computations:

(position of letter: # of choices) : (1st : 25) (2nd : 24) (3rd : 23) (4th : 1) (5th : 22) (6th : 21) (7th : 20) -> 25 x 24 x 23 x 1 x 22 x 21 x 20 = 127,512,000

Since the passwords consist of 7 "distinct" letters, repetition is not allowed. The 4th letter is the middle part of the password that's why there's only one choice there (corresponding to A). The 1st letter has 25 choices (26 less 1 because A is already assigned to the 4th letter). As Prof Mc Gonagall fills the remaining places, the choices decrease by 1.

3. In how many ways can 11 Slytherins line up outside the Divination classroom if Malfoy must be immediately between Crabbe and Goyle?

Answer: 725,760

Here is a representation of the scenario in problem # 3:

S S S S S S S S(CMG)

The S's represent the Slytherin students, C, M and G correspond to Crabbe, Malfoy and Goyle, respectively. Crabbe, Malfoy and Goyle are enclosed in parentheses to denote that they are a group.

To find the number of possible arrangements, we first arrange the 8 Slytherins plus the group that comprises Malfoy and his 2 cronies. So, 8 + 1 group equals 9, therefore there are 9! ways of arranging the Slytherins and Malfoy's group. Next, we look at the group, we can see that there are 2! ways of arranging the people in the group (not 3!) because Malfoy has to stay in between his cronies (what's new?). Therefore, only Crabbe and Goyle can move and there are 2! ways of doing that.

By the fundamental principle of counting, we get the answer to problem # 3 by multiplying 9! with 2!

4. At the Yule Ball during Harry's 4th year, eight students (Hermione, Ron, Fred, George, Neville, and 3 Beauxbatons students) are seated around a table. In how many ways can they be arranged if the 3 Beauxbatons students insist on sitting next to each other and Hermione and a seemingly jealous Ron insist on sitting apart?

Answer: 432

Answer: 3!3!P(4,2) = 432 ways.

First, we arrange the 3 Beauxbatons students(as a group), Fred, George and Neville around the table. Ron and Hermione are not included in this arrangement. If we count the total number of elements around the table, you'll come up with four - 3 students plus the group comprising the Beauxbatons students (3 + 1). Again Hermione and Ron are not included in this arrangement. The total number of ways of arranging the 4 elements are 3! (note: circular permutation -> (n-1)!). Next, we arrange the group of Beauxbatons students. There are 3! ways of arranging them. Finally, we "insert" Ron and Hermione in between the 4 elements. There are 4 spaces in between the 4 elements. Since we only need two, we get the total number of arranging Ron and Hermione by this expression: P(4,2). By doing this, we avoid the two from sitting beside each other because there will be either a student or a group between them.

By the fundamental principle of counting, we multiply 3! by 3! by P(4,2)to get the final answer.

5. In Charms, Harry was asked to use a hovering charm to transfer two blue boxes, three red boxes, five brown boxes, and four yellow boxes onto a shelf. Assuming that boxes of the same color are indistinguishable, in how many ways can they be arranged on the shelf?

Answer: 2,522,520

This is an example of distinct permutations (or constrained repetition). Following this formula:

n!/(n1!n2!n3!...nk!) wherein n1 + n2 + n3 + ... + nk = n

(note: n1 is read as "n sub 1," n2 is "n sub 2," so on until "n sub k")

we obtain: 14!/(2!3!5!4!) = 2,522,520 ways

6. Mrs. Figg invited Ron, Harry, and Hermione to dinner. If she has 10 plates, 7 spoons, 8 forks, and 6 knives, in how many ways can she use them to set the table?

Answer: 7,717,500

Since the order of choosing utensils and plates is not important, we can use combination to solve the problem. Since a dinner set consists of a plate, a spoon, a fork,and a knife; and there are 4 people having dinner, we take 4 of each utensil and 4 plates to set dinner. The total number of ways to do this is:

C(10,4) x C(7,4) x C(8,4) x C(6,4) = 7,717,500

7. A committee of 7 members is to be formed to put up the Christmas decorations at Hogwarts. The following volunteered: 9 teachers, 7 Hufflepuff students, 4 ghosts, 6 Gryffindors, and 3 Prefects. How many possible committees can be formed by Professor Dumbledore if a committee must have only 1 Prefect, only 1 ghost, and at most 2 teachers?

Answer: 216,216

Again, the order of choosing the committee members is not important therefore we use combinations to answer this problem. The term "at most" means that the committee formed can have a maximum of two teachers. This also means that the committee may have 0 teachers, 1 teacher, or 2 teachers. That's why we have to take cases to solve this problem.

Case 1 : 0 teachers -> C(3,1) x C(4,1) x C(9,0) x C(13,5) = 15,444.
Case 2: 1 teacher -> C(3,1) x C(4,1) x C(9,1) x C(13,4) = 77,220.
Case 3: 2 teachers -> C(3,1) x C(4,1) x C(9,2) x C(13,3) = 123,552.
Answer: 15,444 + 77,220 + 123,552 = 216,216 ways.

8. A manufacturing machine in Grunnings produces 100 pieces of drills in 30 minutes. After 1 hour, the drills are divided into groups of 4 and placed in a box. If 2% of the manufactured drills are defective, in how many ways can one select a box containing 2 defective products?

Answer: 114,660

After 1 hour, 200 drills would have been produced. If 2% of the drills are defective, then there are 196 working drills and 4 defective ones. To answer the problem, we will use combinations since the order of choosing the drills is not important. First, we choose the working drills: C(196,2), then we choose the defective ones C(4,2), we multiply the two expressions (by virtue of the fundamental counting principle) to get the final answer:
C(196,2) x C(4,2) = 19,110 x 6 = 114,660 ways.

A box contains 4 drills, and we are looking for one that contains two defective drills. We get the 2 defective drills from the set of 4 defective drills (hence C(4,2)) and we get the working drills from the 196 working drills (hence C(196,2).

9. Dudley encontered this problem in his Math test (It wouldn't be surprising if he failed the test): "What is the 7th term in the expansion of (2x-7)^12?" Find the answer to this problem.

Answer: 6,957,291,264 x^6

Using the formula: C(n,k) * [x^(n-k)] * y^k , we substitute 12 for n, and 6 for k (note: the first term of the expansion has k=0, the 2nd k=1, and so on), 2x for x and -7 for y. We obtain C(12,6) (2x)^(12-6) (-7)^6 = 924(64x^6)(117649) = 6,957,291,264x^6

10. Hermione decided to make codewords for the DA that are composed of 2 letters, and 4 numbers. How many codewords are possible if the letters are taken from the English alphabet and the numbers from 0-9, and repetition is allowed?

Answer: 101,400,000

Answer: C(6,2) x 26 x 26 x 10 x 10 x 10 x 10 = 101,400,000 words.

The answer to this problem seems to be very straightforward. We just simply have to multiply the number of choices for each of the two letters and each of the 4 numbers. However, the problem did not specify how the numbers and letters are placed in the codewords(the problem just mentioned that the codewords comprised of 2 letters and 4 numbers and not how the letters and numbers are arranged in the codeword). Hence, the presence of C(6,2). This represents the choosing of the location (out of 6 locations -> 4 + 2) of the two letters. You can also write C(6,4) if you wish to look for the location of the numbers instead of the letters. The two expressions are equal.

Source: Author metamorphmagus

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A Harry Potter - Themed Quiz on Combinatorics

This quiz is about combinatorics with a Harry Potter theme. It covers permutations, combinations, & their applications. All questions require computations. Arm yourself with a calculator! Enjoy! Don't forget to rate this quiz.

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