Let b₁ b₂ b₃ b₄ be a 4-element permutation with $b_i\in \left\{1,2,3,....,100\right\}for1\le i\le 4and{b}_i\ne b_j$ for $i\ne j,$ such that either b_1, b₂, b₃ are consecutive integers or b₂, b₃, b₄ are consecutive integers. Then the number of such permutations b₁ b₂ b₃ b₄ is equal to……………

Last two digit must be in form

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

Total number of required number = 12 + 18 = 30

Kindly consider the following figure

Sum of digits

1 + 2 + 3 + 5 + 6 + 7 = 24

So, either 3 or 6 rejected at a time

Case 1 Last digit is 2

……….2

no. of cases = ²C₁ × 4! = 48

Case 2 Last digit is 6

……….6

= 4! = 24

Total cases = 72

 $A=\left[\begin{array}{ccc}\hfill a_1\hfill & \hfill a_2\hfill & \hfill a_3\hfill \\ \hfill b_1\hfill & \hfill b_2\hfill & \hfill b_3\hfill \\ \hfill c_1\hfill & \hfill c_2\hfill & \hfill c_3\hfill \end{array}\right],a_2,b_2,c_2\in \left\{0,1\right\}$

S = $a_1+a_2+a_3+b_1+b_2+b_3+c_1+c_2+c_3$ is prime

$0\le s\le 9$

Prime value = 2,3, 5, 7

$ForS=2,1+1+0+0+0+0+0+0+0\to \frac{9!}{2!7!}=36$

Total number of matrices

= 36 + 84 + 126 + 36 = 282

We have,  $1-$  (probability of all shots result in failure)  $>\frac{1}{4}$

$\begin{array}{rr}& \Rightarrow 1-{\left(\frac{9}{10}\right)}^{\mathrm{n}}>\frac{1}{4}\\ & \Rightarrow \frac{3}{4}>{\left(\frac{9}{10}\right)}^{\mathrm{n}}\Rightarrow \mathrm{n}\ge 3\end{array}$