Feedback on Practice Exercises

[WanderingMind]

Hello,
I am just looking for some feedback (whether my answers and reasoning is correct) on some practice exercises I am doing. The subject is currently Combinatorics (didn't see a category for that so i just placed this in general math).

Problem 1:

There are 18 mathematics majors and 325 computer science majors at a college.
a) In how many ways can two representatives be picked so that one is a mathematics major and the other is a computer science major?
My answer is 325!/307!

b) In how many ways can one representative be picked who is either a mathematics major or a computer science major?
My answer is 343

Problem 2:

An office building contains 27 floors and has 37 offices on each floor. How many offices are in the building?
My answer is 27³⁷.

Problem 3:

A multiple choice test contains 10 questions. There are four possible answers for each question.
a) In how many ways can a student answer the questions on the test if the student answers every question?
My answer is 4¹⁰.

b) In how many ways can a student answer the questions on the test if the student can leave answers blank?
My answer is 2¹⁰ * 4¹⁰

Problem 4:

A particular brand of shirt comes in 12 colors, has a male and a female version, and comes in three sizes for each sex. How many different types of this shirt are made?
My answer is 72

 

[berkeman]

Welcome to the PF.

Your answer to #2 does not look correct to me. Can you show your reasoning?

 

[RUber]

1a seems wrong to me. You are essentially asking how many ways can you pick one CS major and one Math major. This should be pretty straightforward. For each choice of one mathematician, you have x number of choices for CS major. If you have y mathematicians, you have xy combinations. If AB is not the same as BA, meaning that order matters, then you would multiply by the number of ways to order 2 items. In this case, the question sounds like order does not matter.

Problem 2, try drawing a picture. Do you have any idea how big 27^37 is? The problem describes a realistic building layout.

3b- Why 2^10 * 4^10? Are there 4 ways to leave the problem blank too? Blank should be seen as a 5th option for response.

The rest look right.

 

[Ray Vickson]

Hello,
I am just looking for some feedback (whether my answers and reasoning is correct) on some practice exercises I am doing. The subject is currently Combinatorics (didn't see a category for that so i just placed this in general math).

Problem 1:

There are 18 mathematics majors and 325 computer science majors at a college.
a) In how many ways can two representatives be picked so that one is a mathematics major and the other is a computer science major?
My answer is 325!/307!

b) In how many ways can one representative be picked who is either a mathematics major or a computer science major?
My answer is 343

Problem 2:

An office building contains 27 floors and has 37 offices on each floor. How many offices are in the building?
My answer is 27³⁷.

Problem 3:

A multiple choice test contains 10 questions. There are four possible answers for each question.
a) In how many ways can a student answer the questions on the test if the student answers every question?
My answer is 4¹⁰.

b) In how many ways can a student answer the questions on the test if the student can leave answers blank?
My answer is 2¹⁰ * 4¹⁰

Problem 4:

A particular brand of shirt comes in 12 colors, has a male and a female version, and comes in three sizes for each sex. How many different types of this shirt are made?
My answer is 72

For problem 1: your answer for (a) is a bit too large:
you get 1013156035751856671370085834080581020876800000, but the actual answer is less than 10000 (exact value known to me). Your answer to (b) is OK.

For problem 2: I cannot figure out how you get $27^{37} =$ 91297581665113611259115979754590511595360241199911147. I get between 800 and 1800---I'll let you work out the exact answer.

For problem (3) (a): your answer is OK.

For problem (4): your answer is OK.

 

[RUber]

3b is not okay.
2^10*4^10 = 8^10, implying that there would be 8 options for each of the 10 questions.

To really overcomplicate the problem, you could say that the number of ways to respond is equal to:
$\sum_{i=0}^{10} \begin{pmatrix} 10\\i \end{pmatrix} 4^i$ indicating that for each i being the number of questions answered, you have 4 ways to answer the question. Summing up all those options is messy.

q's answered: # ways to answer
0 questions: 1
1 question: 10*4
2 questions: 45*16
...
9 questions: 45*4^9
10 questions: 4^10

However, you might recall the binomial expansion
$(a + b) ^ n = \sum_{i=0}^n \begin{pmatrix} n\\i \end{pmatrix} a^i b^{n-i}$.
If you let b = 1, a = 4, and n = 10, you will get the same sum formula I proposed above.

 

[Ray Vickson]

3b is not okay.
2^10*4^10 = 8^10, implying that there would be 8 options for each of the 10 questions.

To really overcomplicate the problem, you could say that the number of ways to respond is equal to:
$\sum_{i=0}^{10} \begin{pmatrix} 10\\i \end{pmatrix} 4^i$ indicating that for each i being the number of questions answered, you have 4 ways to answer the question. Summing up all those options is messy.

q's answered: # ways to answer
0 questions: 1
1 question: 10*4
2 questions: 45*16
...
9 questions: 45*4^9
10 questions: 4^10

However, you might recall the binomial expansion
$(a + b) ^ n = \sum_{i=0}^n \begin{pmatrix} n\\i \end{pmatrix} a^i b^{n-i}$.
If you let b = 1, a = 4, and n = 10, you will get the same sum formula I proposed above.

If a question can be left blank, each question has 5 "answers"---actual answers 1--4 and 'blank'. So, is it not just like 3(a) but with 5 instead of 4?

 

[RUber]

If a question can be left blank, each question has 5 "answers"---actual answers 1--4 and 'blank'. So, is it not just like 3(a) but with 5 instead of 4?

That was the point I was trying to make.

 

[WanderingMind]

Welcome to the PF.

Your answer to #2 does not look correct to me. Can you show your reasoning?

My reasoning was that each floor contains 36 offices. I clearly used the wrong notation here, I should have taken the product of the numbers.

 

[WanderingMind]

Could you clarify the solution to 1a please. I admit that factorials do confuse me at the moment.

 

[RUber]

For 1a, you have 325 CS majors and 18 Math majors.
The question asks how many ways can you choose two people such that 1 is a CS major and 1 is a math major.
I am assuming that order doesn't matter.
There are 325 options for CS majors and 18 options for Math majors.
That is to say, for each of the 325 CS majors, there are 18 possible Math major partners.
There are no factorials needed in this problem. This is simple multiplication.

325! is the total number of different ways you can arrange all 325 CS majors.
325!/307! would be the total number of ways you could arrange 18 CS majors.
And
325!/( 307! 18!) is the total number of ways you could choose 18 CS majors from the pool of 325, when order of selection does not matter.

 

[WanderingMind]

For 1a, you have 325 CS majors and 18 Math majors.
The question asks how many ways can you choose two people such that 1 is a CS major and 1 is a math major.
I am assuming that order doesn't matter.
There are 325 options for CS majors and 18 options for Math majors.
That is to say, for each of the 325 CS majors, there are 18 possible Math major partners.
There are no factorials needed in this problem. This is simple multiplication.

325! is the total number of different ways you can arrange all 325 CS majors.
325!/307! would be the total number of ways you could arrange 18 CS majors.
And
325!/( 307! 18!) is the total number of ways you could choose 18 CS majors from the pool of 325, when order of selection does not matter.

Thank you, now I get it