Permutations and Combinations

Let x = correct answer, y = incorrect answer

  $\therefore 3x-2y=5,x+y\le 5,x,y\in w$              

 only possible (x, y) is (3, 2)

$\therefore$ Required number of ways = 

  $\frac{sin\theta sin\theta }{cos\theta }+\frac{sin\theta }{cos\theta }=2sin\theta cos\theta$

$\Rightarrow \frac{sin\theta }{cos\theta }\left[sin\theta +1\right]=2sin\theta cos\theta$            

$\therefore$ sinθ = 0 and 1 + sin θ = 2 cos² θ = 2 – 2 sin² θ ………….(i)

θ = np

θ = -p, p, 0

From (i), 2 sin² θ + sin θ - 1 = 0

(2 sin θ - 1) (sin θ + 1) = 0

sin θ = -1, $\frac{1}{2}$  

$\theta =\frac{3\pi }{2},\theta =\frac{\pi }{6},\frac{5\pi }{6}$           

$\theta =\frac{3\pi }{2}$ is rejected.

T = cos (-2p) + cos 2p + cos θ + cos $\frac{\pi }{3}+cos\frac{5\pi }{3}=4$  

$\therefore$ T + n(s) = 4 + 5 = 9

Sum of all given numbers = 31

Difference between odd and even positions must be 0,11 or 22 but 0 and 22 are not possible.

$\therefore$ Hence 11 is possible.

This is possible only when either 1, 2, 3, 4 if filled in odd places in order and remaining in other order.

Hence 2, 3, 5 or 7, 2, 1 or 4, 5, 1 at even places.

$\therefore$ Total possible ways = (4! * 3!) * 4 = 576

Fix the unit place, find the chances for the first three digits

unit digit as 1, total ways = 9.10²

unit digit as 2, total ways = 4.5²

unit digit as 3 total ways = 3.4²

unit digit as 4 total ways = 2.3²

unit digit as 5 total ways = 1.2²

unit digit as 6 total ways = 1.2²

unit digit as 7 total ways = 1.2²

unit digit as 8 total ways = 1.2²

unit digit as 9 total ways = 1.2²

 $f\left(x\right)=2co{s}^{-1}x+4co{t}^{-1}x-3x^2-2x+10\forall x\in \left[-1,1\right]$

$\Rightarrow f\text{'}\left(x\right)=-\frac{2}{\sqrt[]{1-x^2}}-\frac{4}{1+x^2}-6x-2<0\forall x\in \left[-1,1\right]$

So, f (x) is decreasing function and range of f (x) is

$\left[f\left(1\right),f\left(-1\right)\right],$ which is $\left[\pi +5,5\pi +9\right]$

Now 4a – b = 4 ( + 5) (5 + 9) = 11 - π

$co{s}^{-1}\left(\frac{3}{10}cos\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)+\frac{2}{5}sin\left(ta{n}^{-1}\left(\frac{4}{3}\right)\right)\right)$

$=co{s}^{-1}\left(\frac{3}{10}.\frac{3}{5}+\frac{2}{5}.\frac{4}{5}\right)$

$=co{s}^{-1}\left(\frac{1}{2}\right)=\frac{\pi }{3}$

Three consecutive integers belong to 98 sets and four consecutive integers belongs to 97 sets.

Þ Number of permutations of b₁ b₂ b₃ b₄ = number of permutations when b₁ b₂ b₃ are consecutive + number of permutations when b₂, b₃, b₄ are consecutive – b₁ b₂ b₃ b₄ are consecutive = 98 * 97 * 98 * 97 – 97 = 18915

Last two digit must be in form

$\begin{array}{l}23,4,516\\ 324\\ 32\\ 52\end{array}\}3*4=12$

Total number of required number = 12 + 18 = 30

Kindly consider the following figure