Prerequisite : Permutation and Combination

**n students appear in an examination, find the number of ways the result of examination can be announced. **

Answer is

2^{n}Examples:

Input : n = 6

Output : Each student can either pass or fail in the examination. so ,there exists 2

possibilities for each of the 6 students in the result. hence total number of ways for the result=(2)^{6}Input : n = 8

Output :(2)^{8}=256

**‘n’ matches are to be played in class a chess tournament, find the number of ways in which their results can be decided**

Answer is (3)

^{n}waysExamples:

Input : n = 3

Output: The results of each of the 3 matches can be three ways namely win ,draw or loss

since total no. of ways in which results of 3 matches can be decided =(3)^{3}Input:

6

Output:(3)^{4}=81

**A badminton tournament consists of ‘n’ matches.
(i) Find the number of ways in which their results can be forecast are given.
(ii) Total number of forecasts containing all correct results.
(ii) Total number of forecasts containing all wrong results.**

Answer (i) (2

^{n})

(ii) 1

(iii) 1

Examples:

Input : A badminton tournament consists of 3 matches.

(i) In how many ways can their results be forecast ?

(ii) How many different forecasts can contain all correct results ?

(iii) How many different forecasts can contain all correct results ?

Output:(i) Each badminton match can be decided in only 2 ways either win or

loss for a particular team so total number of ways results of 3

matches can be forecast=2^{3}=8

(ii) Results of each match can be forecast wrong in only 1 way

Total no. of forecasts containing all wrong results = (1^{3}) = 1

(iii) Similarly, result of each can be forecast correct in only 1 way.

total no .of forecasts containing all correct results = (1^{3}) = 1

** Find the number of ways in which ‘n’ different beads can be arranged to form a necklace**

Answer is (n-1)!/2

Examples: For example 4 beads can be arranged in following ways.

….b1b2…….b4

….b3

….b1

b3…….b2

….b4

….b1

b4…….b3

….b2

Since it does not matter where we place first bead. Total ways to arrange is (n – 1)!. But clockwise and anticlockwise arrangements are same, so total arrangements are (n – 1)!/2

** There are ‘n’ questions papers, find the no, of ways in which a student can attempt one or more questions**

Answer:

(2^{n}-1)ways.For example a student will solve one or more questions out of 4 questions in following ways.

1) The student chooses to solve only one question, can choose in^{4}C_{1}

2) The student chooses to solve only two questions, can choose in^{4}C_{2}

3) The student chooses to solve only three questions, can choose in^{4}C_{3}

3) The student chooses to solve all four questions, can choose in^{4}C_{4}

So total ways are

^{4}C_{1}+^{4}C_{2}+^{4}C_{3}+^{4}C_{4}

=2^{4}-1 = 15 ways

We know sum of binomial coefficients from^{n}C_{0}to^{n}C_{n}is 2^{n}

**More practice questions on permutation and combination : **

Quiz on Permutation and Combination

Combination and Permutation Practice Questions

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