Permutations and Combinations of answers and questions

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Homework Help Overview

The discussion revolves around combinatorial problems, specifically focusing on permutations and combinations. The original poster presents a scenario involving a candidate who must select questions from two sections of an exam, with specific constraints on the selection process. Additional problems regarding the arrangement of balls in boxes and the formation of three-digit numbers are also introduced.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants express confusion regarding the original problem due to missing information about the options in section B. Some participants suggest that without this information, it is challenging to determine the number of combinations.
  • In subsequent posts, participants attempt to solve related problems involving indistinguishable and distinguishable objects, discussing various combinatorial methods.

Discussion Status

The discussion is ongoing, with participants providing insights into different problems. Some have attempted to clarify the original question, while others have begun to offer solutions to the additional problems presented. There is no explicit consensus on the original question's approach, but several lines of reasoning are being explored.

Contextual Notes

Participants note the constraints of the original problem, particularly the lack of information regarding the number of questions in each option of section B. This uncertainty is acknowledged as a significant barrier to finding a solution.

look416
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Homework Statement



A candidate sitting this paper is told to answer 5 of the 7 questions in section A, and 3 questions from the 5 options in section B, where not more than 2 questions from the same option can be chosen. Assuming that he answers 8 questions altogether, find how many different combinations of questions he could attempt.

Homework Equations



nPr , nCr

The Attempt at a Solution



To speak the truth, i can't solve this questions
even my teacher also don't know how to solve it =.=
but i got the answer, the answer is 9450,
just want to get the approach only...
 
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look416 said:

Homework Statement



A candidate sitting this paper is told to answer 5 of the 7 questions in section A, and 3 questions from the 5 options in section B, where not more than 2 questions from the same option can be chosen. Assuming that he answers 8 questions altogether, find how many different combinations of questions he could attempt.

Homework Equations



nPr , nCr

The Attempt at a Solution



To speak the truth, i can't solve this questions
even my teacher also don't know how to solve it =.=
but i got the answer, the answer is 9450,
just want to get the approach only...
Without knowing how many questions there are in each "option" of B, I don't see how you can answer this.
 
HallsofIvy said:
Without knowing how many questions there are in each "option" of B, I don't see how you can answer this.

thanks for informing me the question's problem
 
another 2 questions

2. Three balls are to be placed in three different boxes, not necessarily with one ball in each box. Any box can hold one, two or three balls. Find the number of ways the ball can be replaced.
a. if they are all of the same colour and therefore indistinguishable
b. if they are all of different colours.

3. Find how many three figure numbers, lying between 100 and 999 inclusive, have two and only two consecutive figures identical.
 
Well, you should try something and write your work. If you are stuck, someone will help you, but you have not shown any work yet.
 
your question 1 should be written as:

A candidate sitting this paper is told to answer 5 of the 7 questions in section A, and 3 questions from the 5 options(each option consists of 3 questions) in section B, where not more than 2 questions from the same option can be chosen. Assuming that he answers 8 questions altogether, find how many different combinations of questions he could attempt.

later i can give you the solutions to questions 2 and 3 which are quite easy
 
Answer to question 2a.

If the balls are of the same colour there is only 3C1 = 3 ways in which all of the 3 balls should be in one box.
and there is only 1 way in which 1 ball is in each box (they are of the same colour)
and there are only 3C1 X 2C1 = 6 ways of placing one ball in any of the 3 boxes and 2 and the remaining two boxes.

hence total no of possibilities = 1+3+6 =10
 
Answer to question 2b.

if the balls are of different colour then there are 3! =6 ways in which they can be placed 1 in each box

there are also 3C1 =3 ways in which all the balls go in one box

and finally there are 3C1 X 3C2 X 2C1 =18 ways in which we can choose one box containing 2 ball and the remaining 1 ball will go into any of the other 2 boxes.

hence total no of possibilities = 3+6+18 =27
 
Answer to question 3.

just think a bit and u will get the answer as = 81 +81 =162

(Hint 100, 122, 133, 144,...,199 and same goes for the other 9 figures : 1,2,3, ...,9)

then (110, 112,113,114,...,119 and same goes for the other 9 figures : 1,2,3, ...,9)
 

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