Fundamental principles of counting - class-XI

According to question,

For any 5 points chosen from 6 equally spaced points on a circle, the area of the resulting convex pentagon is determined by the relative positions of the points. Due to the symmetry of the points on the circle, all convex pentagons formed by choosing 5 out of 6 points are congruent and thus have the same area.

Selection of$ 4$ digits out of$ 10$ (including 0)=$ ^{10}C _4$

The order pair $(x, y)$ satisfying $xy=225$ are $(1, 225), (3, 75) (5, 45), (9, 25), (15, 15)$. Order can be changed in the first four pairs and both $x$ and $y$ can be negative also, so the no. of pairs $=2(2\times 4+1)=18$

(A) Since every edge connects a pair of points, the given 12 points have to be joined using lines. We may have minimum number of edges if all the 12 points are collinear.
No. of edges in this particular case 
$=12-1=11$
Maximum number of edges are possible when all the 12 points are non-collinear. In this particular case number of different straight lines that can be formed using 12 points which is equal to $^12C _{2}$
$=\frac{12\times 11}{2}=66$
Therefore, following inequality holds for "e"
$11 \leq e  \leq 66$

Numbers 1 to 90 can be classified by remainder when divided by 3: 30 numbers each in categories congruent to 0, 1, and 2 mod 3. For sum of two numbers to be divisible by 3: either both are ≡ 0 (mod 3), or one is ≡ 1 and other ≡ 2 (mod 3). Ways from first category: C(30,2) = 30×29/2 = 435. Ways from second category: 30 × 30 = 900. Total ways = 435 + 900 = 1335.

(B) The number of ways in which 1 green ball can be put $=6$ . The number of ways in which two green balls can be put such that the boxes are consecutive 
$=5$ $(i.e., (1, 2),(2, 3),(3, 4),(4, 5),(5, 6))$

Similarly, the number of ways in which three green balls can be put 
$=4( i.e. (1, 2, 3),(2, 3, 4),(3, 4, 5),(4, 5, 6))$
$\cdots \cdots \cdots \cdots \cdots $ and so on.
$\therefore $ Total number of ways of doing this
$=6+5+4+3+2+1=21$

For a particular class, the total number of different tickets from first intermediate station is $5.$ 
Similarly, number of different tickets from second intermediate station is $4.$ 
So the total number of different tickets is $5+4+3+2+1=15$.
And same number of tickets for another class is equal to total number of different tickets, 

Matches played between 8 teams =$ 7+6+5+4+3+2+1 = 21$

The total number of n-digit numbers using digits 1 and 2 is 2^n. Subtracting the cases where either 1 or 2 is never used gives 2^n - 2 numbers where both digits appear at least once. Setting 2^n - 2 = 510 gives 2^n = 512, which means n = 9.

Total number of wrong answers $= 2^{n-1}+2^{n-2}+...+2^{2}+2+1$
$=2^{n-1}+2^{n-2}+...+2^{4}+2^{3}+2^{2}+2+1$
$\Rightarrow 2^{n}-1=2043$  [Using formula for sum of G.P]
$ \Rightarrow 2^{n}=2048$
$ \Rightarrow 2^{n}=2^{11}$
$\Rightarrow n = 11$
Hence, option 'B' is correct.

In order to solve this question, we must observe the number of ways in which we can select the first term of the required A.P. for different values of the common difference($r$) starting from $r=1$ given that there are only $3$ terms before $n$.
For $r=1$, the number of ways in which we can select the first term of the A.P. $=n-2$
For $r=2$, the number of ways to select the first term $=n-4$
For $r=3$, the number of ways to select the first term $=n-6$
Now we see a pattern emerging, we also realize from this that $r<=\dfrac{n-1}{2}$ for an A.P. with 3 terms to exist in the given interval.
$\therefore$ The final answer $=n-2+n-4+n-6+...+5+3+1$
Now we use the formula to find the sum of an A.P. which is $S _n=\dfrac{n}{2}[a _1+a _n]$ 
$\therefore$ Answer $=\dfrac{n-1}{4}[1+n-2]=\dfrac{(n-1)^2}{4}$

$7$ Courses in morning 

We know that $ { { n } _{ C } } _{ r }=  { { n } _{ C } } _{n- r }$

Given, $ { { n } _{ C } } _{3}= { { n } _{ C } } _{13} $
$ => r = 3 $ and $ n -r = 13 $
$ => n - 3 = 13 $
$ => n = 16 $

Also, $ { { n } _{ C } } _{ r }=\dfrac { n! }{ r!(n-r)! }  $
So, $ { { 20 } _{ C } } _{ 16 }=\dfrac { 20! }{ 16!(20-16)! } = \dfrac { 20! }{ 16! \times 4! }  = \dfrac { 20 \times 19 \times 18 \times 17 \times 16! }{ 16! \times 4 \times 3 \times 2 } = 4845   $

If two zeros are the entries in the diagonal, then
$^{3}\mathrm{C} _{2}\times^{3}\mathrm{C} _{1}$
If all the entries in the principle diagonal is 1, then
$^{3}\mathrm{C} _{1}$
$\Rightarrow$  Total matrix $= 12.$

Clearly, every member except one (i.e. the winner) must lose one game to decide the winner. Thus, minimum number of matches to be played = 30 - 1 = 29.