Combinatorics MCQ Quiz in मल्याळम - Objective Question with Answer for Combinatorics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക
Last updated on Mar 15, 2025
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Combinatorics Question 1:
Let \(a_{n}\) denote the number of all \(n\)-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them are 0. Let \(b_{n}=\) the number of such \(n\)-digit integers ending with digit 1 and \(c_{n}=\) the number of such \(n\)-digit integers ending with digit 0. The value of \(b_{6}\) is
Answer (Detailed Solution Below)
Combinatorics Question 1 Detailed Solution
1. 1 0 1 1 1 1
2. 1 0 1 0 1 1
3. 1 1 1 1 1 1
Three cases possible:
1. One zero- It can be placed in any of the four places. So, we get '4' such six digit numbers.
2. Two zeros- We get '3' such six digit numbers possible.
3. No zeros- We get only '1' such six digit number.
Hence \(b_6=4+3+1=8\)
Combinatorics Question 2:
\( { a }_{ 12 } \)
Answer (Detailed Solution Below)
Combinatorics Question 2 Detailed Solution
\( a_{n+2}-a_{n+1}-a_n=p\alpha^n(\alpha^2-\alpha-1)+q\beta^n(\beta^2-\beta-1)=0 \)
So,
\( a_{12}-a_{11}-a_{10}=0 \)
\( a_{12}=a_{11}+a_{10} \)
Combinatorics Question 3:
What is the minimum number of students needed in a class to guarantee that there are at least 6 students whose birthdays fall in the same month?
Answer (Detailed Solution Below)
Combinatorics Question 3 Detailed Solution
Key Points
Pigeon hole principle:
In general, if K is a positive integer and KN+1 pigeons are distributed among 'n; pigeon holes then some hole contains at least K+1 pigeons.
This problem is the same as the above concept, the minimum number of students needed in a class to guarantee that there are at least 6 students whose birthdays fall in the same month over a year. A year has 12 months So,
\(12 \times 6 +1= 61\)
The minimum number of students are=61.
Hence the correct answer is 61.
Combinatorics Question 4:
What is the maximum number of regions that the plane R2 can be partitioned into using 10 lines?
Answer (Detailed Solution Below)
Combinatorics Question 4 Detailed Solution
Key Points
The recurrence is given by A(n)=A(n−1)+n. Each new nth line drawn is creating n new partitions. While creating partitions, draw the new line in such a way that it cuts all the previously drawn n−1 lines, then the nth line will create n new partitions and previous A(n−1) partitions will remain the same.
A(n)=A(n−1)+n. A(0)=1,A(1)=2,A(2)=4
A(10)=A(9)+10=46+10=56
A(9)=A(8)+9=37+9=46
A(8)=A(7)+8=29+8=37
A(7)=A(6)+7=22+7=29
A(6)=A(5)+6=16+6=22
A(5)=A(4)+5=11+5=16
A(4)=A(3)+4=7+4=11
A(3)=A(2)+3=4+3=7
Hence the correct answer is 56.