Permutations and Combinations of answers and questions

In summary, the candidate was instructed to answer 5 of the 7 questions in section A and 3 questions from the 5 options in section B, with a maximum of 2 questions from the same option. The total number of combinations of questions that could be attempted is 9450. For the second question, there are 10 possibilities if the balls are of the same color and 27 possibilities if the balls are of different colors. For the third question, there are 162 three-digit numbers between 100 and 999 inclusive that have two consecutive identical digits.
  • #1
look416
87
0

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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  • #2
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.
 
  • #3
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
 
  • #4
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.
 
  • #5
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.
 
  • #6
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
 
  • #7
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
 
  • #8
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
 
  • #9
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)
 

What is the difference between permutations and combinations?

Permutations refer to the different ways in which a set of items can be ordered or arranged, while combinations refer to the different ways in which a subset of items can be selected from a larger set, without regard to order.

How do you calculate the number of permutations?

The number of permutations can be calculated using the formula nPr = n! / (n-r)!, where n is the total number of items and r is the number of items being selected.

How do you calculate the number of combinations?

The number of combinations can be calculated using the formula nCr = n! / (r!(n-r)!), where n is the total number of items and r is the number of items being selected.

Can you give an example of permutations and combinations in real life?

An example of permutations in real life would be the different ways in which a group of people can be arranged in a line. An example of combinations would be the different ways in which a group of people can be selected to form a team.

How can permutations and combinations be used in probability calculations?

Permutations and combinations are often used in probability calculations to determine the likelihood of a certain outcome occurring. For example, in a lottery with 6 numbers, the number of possible combinations of winning numbers would be calculated using combinations.

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