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Comment on Bicycle with Optional Features
Is this FCP problem or
So there are such type of questions that can be solved using both the methods?
It can be solved using either
It can be solved using either the Fundamental Counting Principle (FCP) or combinations.
Both approaches are discussed in the video.
From 0:45 to 1:35, I use the FCP approach.
From 1:35 to 3:05, I use combinations.
Could you please help me out
In an entrance test a candidate is required to attempt a total of four questions which are to be attempted from 2 section each containing 5 questions.The maximum number of questions that he can attempt from any section is 3.In how many ways can he answer in the test.
For the above example I tried doing 5C3×5C1 + 5C2x5C2 + 5C1x5C3.
I wanted to know whether it is correct to consider 5C0x5C4 also.
Your solution (5C3×5C1 +
Your solution (5C3×5C1 + 5C2x5C2 + 5C1x5C3) already considers all possible cases.
That is, if we call one section Section A and call the other section Section B, then there are 3 possible cases:
ANSWER 3 QUESTIONS FROM SECTION A AND 1 QUESTION FROM SECTION B
Number of possible outcomes = 5C3 x 5C1
ANSWER 2 QUESTIONS FROM SECTION A AND 2 QUESTIONS FROM SECTION B
Number of possible outcomes = 5C2 x 5C2
ANSWER 1 QUESTION FROM SECTION A AND 3 QUESTIONS FROM SECTION B
Number of possible outcomes = 5C1 x 5C3
So, TOTAL outcomes = 5C3×5C1 + 5C2x5C2 + 5C1x5C3
The last value you suggested (5C0x5C4) represents answering 0 questions from Section A and answering 4 questions from Section B. This breaks the rule that says "The maximum number of questions that he can attempt from any section is 3." So, we must not include this in our final answer.
Does that help?
I got the correct answer, but
If I point to a table of 5
If I point to a table of 5 donuts and say that you can have ANY NUMBER of donuts (up to all 5), I think it's clear that you can choose to have zero donuts.
More importantly, there's nothing in the question that says a bicycle must have at least 1 feature
I see your point. However,