Prerequisite : Permutation and Combination
n students appear in an examination, find the number of ways the result of examination can be announced.
Answer is 2n
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
‘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 ways
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
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) (2n)
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=23=8
(ii) Results of each match can be forecast wrong in only 1 way
Total no. of forecasts containing all wrong results = (13) = 1
(iii) Similarly, result of each can be forecast correct in only 1 way.
total no .of forecasts containing all correct results = (13) = 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.
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
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 4C1
2) The student chooses to solve only two questions, can choose in 4C2
3) The student chooses to solve only three questions, can choose in 4C3
3) The student chooses to solve all four questions, can choose in 4C4
So total ways are
4C1 + 4C2 + 4C3 +4C4
=24-1 = 15 ways
We know sum of binomial coefficients from nC0 to nCn is 2n
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
- Number of Symmetric Relations on a Set
- Mathematics | Matrix Introduction
- Find sum of even index binomial coefficients
- Relationship between number of nodes and height of binary tree
- Number of Reflexive Relations on a Set
- Dilworth’s Theorem
- Print all the combinations of N elements by changing sign such that their sum is divisible by M
- Number of ways to form an array with distinct adjacent elements
- Minimum number of operations to convert a given sequence into a Geometric Progression
- Number of permutations such that sum of elements at odd index and even index are equal