Lesson: Introduction to Combinations

Comment on Introduction to Combinations

I could not find part two of the articles.
greenlight-admin's picture

Thanks for the heads up!
I have fixed the broken link.

Cheers,
Brent

Cant understand the difference between the sum in link two and link 1 .both talk about selecting nos from a a set and still the first uses the mississippi rule. Can u please help me out with how the 1st sum uses the mississippi rule
greenlight-admin's picture

Link #1: http://gre.myprepclub.com/forum/set-s-consists-of-5-objects-1900.html

Link #2: http://gre.myprepclub.com/forum/qotd-23-s-2619.html

The solution for question #1 is not applying the Mississippi rule. It's applying the combination formula.
That is, we can choose 4 objects from 5 objects in 5C4 ways.
5C4 = 5!/(4!1!)

The solution for question #2 also applies the combination formula.

Does that help?

Cheers,
Brent

https://gre.myprepclub.com/forum/set-s-consists-of-5-objects-1900.html


Can you please explain this? I did not get your explanation neither do I get Sandy's
greenlight-admin's picture

https://gre.myprepclub.com/forum/main-course-chicken-beef-tofu-side-dish-rice-salad-sou-11831.html

I thought this problem would consist of four slots, one for each step, like picking options for a car.

The first step requires picking one main course out of three choices. Since there are 3 choices I put 3 in the first slot.

Second step requires picking one side. I have 4 choices. I put 4 in the second slot.

Third step requires picking 2nd side. I have 3 choices left (if repeating a side is not allowed).

Fourth step requires picking a desert. I have 2 choices.

If I fill in the four slots I get

3x4x3x2 = 72

Which is wrong.

I guess since the two sides are drawn from the same pool of choices that should indicate to me those two choices should be consolidated into one slot that should be filled by the result of 4C2.

I guess the takeaway is whenever I see multiple choices from a single pool that is not considered different steps in a process, but a single step, like picking committee members. Correct?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/main-course-chicken-beef-tofu-side-dish...

Your analysis is correct; in your solution, you're counting some outcomes twice.

For example, one outcome is: Beef-Rice-Pasta-Cake
And another outcomes is Beef-Pasta-Rice-Cake
However, these two outcomes are the same.
To account for this duplication, we must divide your answer (of 72) by 2.

Cheers,
Brent

Regarding: Combinations and Non-Combinations – Part III (article) https://www.greenlighttestprep.com/articles/combinations-and-non-combinations-%E2%80%93-part-iii

In the sentence, "Does the outcome of each step differ from out outcomes of the other steps?”

I assume the word "out" is a typo.
greenlight-admin's picture

Link: https://www.greenlighttestprep.com/articles/combinations-and-non-combina...

Yes, definitely a typo.
It should have read "Does the outcome of each step differ from the outcomes of the other steps?”
I've now fixed that article.

Thanks for the heads up!!

Could you help me with this question: In an examination, there are four multiple choice questions each with 3 options, out of which one is the correct option. Let K be the number of ways in which questions can be answered such that at least one question is answered correct. Find quantity K
greenlight-admin's picture

Number of ways to answer 1st question = 3
Number of ways to answer 2nd question = 3
Number of ways to answer 3rd question = 3
Number of ways to answer 4th question = 3
Total number of ways to complete the test = (3)(3)(3)(3) = 81

HOWEVER, the correct answer is not 81, because some of those outcomes feature tests in which ZERO questions were answered correctly.

So we must subtract from 81 the total number of ways we can answer 0 questions correctly.
Number of ways to incorrectly answer 1st question = 2
Number of ways to incorrectly answer 2nd question = 2
Number of ways to incorrectly answer 3rd question = 2
Number of ways to incorrectly answer 4th question = 2
Total number of ways to get all 4 answers wrong = (2)(2)(2)(2) = 16

So, the number of ways to complete the test so that AT LEAST one question is correct = 81 - 16 = 65

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