Permutations and Combinations

1^st position can be filled in 4 ways as zero cannot appear in 1^st position.

2^nd position can be filled in 4 ways and so on.

Total cases = 4 * 4 * 3 * 2 * 1 = 96

$ta{n}^{-1}\left(\frac{2*\frac{3}{5}}{1-\frac{9}{25}}\right)+si{n}^{-1}\frac{5}{13}\Rightarrow ta{n}^{-1}\frac{15}{8}+ta{n}^{-1}\frac{5}{12}=ta{n}^{-1}\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\frac{5}{12}}=ta{n}^{-1}\frac{220}{21}$

$\therefore tan\left(ta{n}^{-1}\frac{220}{21}\right)=\frac{220}{21}$

Total 4 digit number

4 digit divisible by 7

1001, 1008, -9996

9996 = 1001 + (n₁ – 1) 7

n₁ = 1286

4 digit no divisible by 3

1002, 1005, -9999

9999 = 1002 + (n₂ – 1)3

n₂ = 3000

4 digit number visible by 21

1008, 1031, -9996

n₃ = 429

4 digit number divisible by 7 or 3

= 9000 – 1286 – 3000 + 429

= 5143

VOWELS

vowels – 2, constants – 4

all the consonants never come together = 6! – 3! 4! =720 – 144 = 576

Let $l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{x}^2}{lo{g}_e{x}^2+lo{g}_e\left(x^2-44x+484\right)}dx..........\left(i\right)}$

By property ${\displaystyle \underset{a}{\overset{b}{\int }}f\left(x\right)dx={\displaystyle \underset{a}{\overset{b}{\int }}f\left(a+b-x\right)dx}}$

$\left(i\right)\Rightarrow l={\displaystyle \underset{6}{\overset{16}{\int }}\frac{lo{g}_e{\left(22-x\right)}^2}{lo{g}_e{\left(22-x\right)}^2+lo{g}_e{x}^2}dx}..........\left(ii\right)$      

(i) + (ii) 2l = ${\displaystyle \underset{6}{\overset{16}{\int }}1dx=10}$

l = 5

Kindly consider the following figure

For divisibility by 5 last digit must be 0 or 5 but 0 is not possible in palindrome

so it will be 5.

So, required no. = 10 * 10 = 100

Total number = 20 + 8 + 24 = 52