Combinatorics MCQ Quiz in मल्याळम - Objective Question with Answer for Combinatorics - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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 $b_6=4+3+1=8$

$a_{n+2}-a_{n+1}-a_n=p{\alpha }^n({\alpha }^2-\alpha -1)+q{\beta }^n({\beta }^2-\beta -1)=0$

So,

$a_{12}-a_{11}-a_{10}=0$

$a_{12}=a_{11}+a_{10}$

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.

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.