Numbers are to be formed between 1000 and 3000, which are divisible by 4, using the digits 1,2,3,4,5 and 6 without repetition of digits. Then the total number of such numbers is………

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

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)  $>\frac{1}{4}$

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