Solving Two Stat Word Problems: Permutations and Combinations Explained

  • Thread starter kuahji
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In summary: A" and "B". There are six different cases. case 1: A and B are both next to each other case 2: A is next to B, but not next to C case 3: B is next to C, but not next to A case 4: A is next to B, but not next to C case 5: B is next to A, but not next to C case 6: A is next to B, but not next to C So in total, there are 6!/(2!*3!*4!*5!*6!) = 120 possibilities.
  • #1
kuahji
394
2
I'm having a bit of difficulty thinking through two problems.

A shipment of 10 television sets includes three that are defective. In how many ways can a hotel purchase four of these sets an receive at least two of the defective sets?

Without making a tree diagram, I'd like to understand the problem, & how to correctly write it. I know the answer is 70, because that is what the book says. I assumed I'd want a permutation, because it appears order will matter. I tried (7!/4!)/(3!/2!) & got 70. But it still doesn't make much sense.

In how many ways can five persons line up to get on a bus? In how many ways can they line up if two of the persons refuse to follow each other?

Here again I'm going to need to use a permutation. So in part one I did 5! & got 120, which was correct, that makes sense. But for part two the answer is 72. Which I believe will be 5!/?, I tried 5!/2! assuming because of the two people that won't follow each other, but that clearly didn't work. Any ideas how to make sense of these two problems?
 
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  • #2
kuahji said:
I'm having a bit of difficulty thinking through two problems.

A shipment of 10 television sets includes three that are defective. In how many ways can a hotel purchase four of these sets an receive at least two of the defective sets?

Well, how many different ways are there to pick 4 televisions out of the ten? There are 210 ways. So, just right down all 210 possibilities and see how many of them contain at least two broken sets.

Without making a tree diagram, I'd like to understand the problem, & how to correctly write it. I know the answer is 70, because that is what the book says. I assumed I'd want a permutation, because it appears order will matter. I tried (7!/4!)/(3!/2!) & got 70. But it still doesn't make much sense.

In how many ways can five persons line up to get on a bus? In how many ways can they line up if two of the persons refuse to follow each other?

Here again I'm going to need to use a permutation. So in part one I did 5! & got 120, which was correct, that makes sense. But for part two the answer is 72. Which I believe will be 5!/?, I tried 5!/2! assuming because of the two people that won't follow each other, but that clearly didn't work. Any ideas how to make sense of these two problems?
 
  • #3
olgranpappy said:
Well, how many different ways are there to pick 4 televisions out of the ten? There are 210 ways. So, just right down all 210 possibilities and see how many of them contain at least two broken sets.

Thats a bit tedious. There must be a better way to think through the problem then trying to write down 210 possibilities. If I use that method, it'll work, but I'll only be able to work 1 problem in the given amount time when exam comes.
 
  • #4
Of course there's an easier way. You need to use combinations here. I'll give you a start.
As mentioned 10C4 = 210 is the total no of ways of picking 4 TVs. Subtract from this the no. of ways of getting all good TVs and the no of ways of getting exactly one bad TV and 3 good TVs and you have your answer. Can you finish it off?

Second question, consider the two not so friendly people as one unit. This should help you find the no of arrangements in which they come together. Subtract this from the total no. of arrangements.
 
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  • #5
This is similar to a hypergeometric probability. Try looking that up in your textbook.
 
  • #6
It is hypergeometric.

Just find the Probability of the occurrence, then multiply that by the total number of ways you can purchase 4 sets to get the E(x)

=P(two defective) + P(three defective) + P(four defective)
=[ 3C2 * 7C2 ] / [ 10C4] + [ 3C3 * 7C1] / [10C4] + 0

So P(x) * n = E(x), which should amount to 70 (I haven't calculated that yet)
 
  • #7
I know that this is way late in responding, but I figured I'd share the way that I solved the bus problem since I just figured it out myself (and stumbled upon this post for help). since there are a total of 120 ways that all 5 can be arranged in the line, we have to look at the different cases of that where the two people are next to each other. Let's refer to the two people who do not want to be next to each other A and B.

For example, you can have:

A B 3 2 1 = 6 choices (1*1*3*2*1 = 6 permutations with person A first and B second)

Repeat this process for A in slots 1-4 and repeat for B in slots 1-4. Subtracting the total number of ways this can happen from the original 120 should give you the answer you need.

Hope this helps.
 
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  • #8
darkmagicianoc said:
I know that this is way late in responding, but I figured I'd share the way that I solved the bus problem since I just figured it out myself (and stumbled upon this post for help). since there are a total of 120 ways that all 5 can be arranged in the line, we have to look at the different cases of that where the two people are next to each other. Let's refer to the two people who do not want to be next to each other A and B.

For example, you can have:

A B 3 2 1 = 6 choices (1*1*3*2*1 = 6 permutations with person A first and B second)

Repeat this process for A in slots 1-4 and repeat for B in slots 1-4. Subtracting the total number of ways this can happen from the original 120 should give you the answer you need.

Hope this helps.

Sometimes people forget to consider the arrangements where B is before A (i.e. BA321).

It is often helpful (as someone mentioned) to think of the two "enemies" as a fused unit (for exclusion purposes) and figure how many ways the 4 objects can be arranged and then multiply this by how many ways the enemies can be arranged within the diad. Subtract this from the original unrestricted orderings and you get your answer.

--Elucidus
 
  • #9
It very easy to solve this problem. Just you need to use combinations.

[tex]C_7^1*C_3^3 + C_7^2*C_3^2=7+63=70[/tex]

:smile:
 

What are "Two Stat Word Problems"?

"Two Stat Word Problems" are a type of mathematical question that involves two different sets of data, or statistics, and requires you to use equations or formulas to find a solution. These types of problems are often used in statistics, economics, and other fields that involve analyzing data.

How do I solve a "Two Stat Word Problem"?

The first step in solving a "Two Stat Word Problem" is to carefully read and understand the problem. Then, identify the two sets of data and determine which equations or formulas are needed to solve the problem. Next, plug in the given values and solve for the unknown variable. Finally, check your answer to ensure it makes sense and is supported by the given data.

What are some common types of "Two Stat Word Problems"?

Some common types of "Two Stat Word Problems" include finding the mean, median, and mode of two sets of data, calculating percentages or ratios based on two sets of data, and analyzing trends or relationships between two variables.

What skills are needed to solve "Two Stat Word Problems"?

To solve "Two Stat Word Problems", you will need a strong understanding of basic mathematical concepts, such as equations, formulas, and statistics. Critical thinking skills are also important, as you will need to be able to analyze and interpret the given information to determine the appropriate solution.

How can I improve my ability to solve "Two Stat Word Problems"?

The best way to improve your ability to solve "Two Stat Word Problems" is to practice regularly. You can find many examples and practice problems online or in math textbooks. It can also be helpful to review basic mathematical concepts and formulas, and to seek assistance from a tutor or teacher if you are struggling with a specific type of problem.

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