# Permutations and Combinations

1. Sep 9, 2009

### look416

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

nPr , nCr

3. The attempt at a solution

To speak the truth, i cant solve this questions
even my teacher also dont know how to solve it =.=
just want to get the approach only....

2. Sep 10, 2009

### HallsofIvy

Staff Emeritus
Without knowing how many questions there are in each "option" of B, I don't see how you can answer this.

3. Sep 10, 2009

### look416

thanks for informing me the question's problem

4. Sep 10, 2009

### look416

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. Sep 10, 2009

### njama

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. Dec 4, 2009

### axlsaml1

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. Dec 4, 2009

### axlsaml1

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. Dec 4, 2009

### axlsaml1

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. Dec 4, 2009