Permutations and Combinations

A → 5Q; B → 5Q; C → 5QA
A?, A?, A?, A?, A?; B?, B?, B?, B?, B?; C?, C?, C?, C?, C?
A?A?A?B?C? → ?C? * ?C? * ?C? = 750
A?A?B?B?C? → ³C? * ?C? * ?C? = 1500
∴ Total = 2250

$\mathrm{}6\cdot {}^{35}{\mathrm{C}}_{\mathrm{r}}=\left({\mathrm{k}}^2-3\right)\cdot \frac{36}{\mathrm{r}+1}{}^{35}{\mathrm{C}}_{\mathrm{r}}$

$\Rightarrow {\mathrm{k}}^2-3=\frac{\mathrm{r}+1}{6},{\mathrm{k}}^2-3>0$

(i) $k=\pm 2$ gives $r=5$

(ii) $k=\pm 3$ gives $r=35$

4 ordered pairs

? P? =? P? ⇒ n!/ (n-r)! = n!/ (n-r-1)! ⇒ n-r=1.
? C? =? C? ⇒ r + (r-1) = n ⇒ n = 2r-1.
Substitute n: (2r-1)-r=1 ⇒ r=2.

I (6)    F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

  ${\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{1}{1+r+r^2}={\displaystyle \sum _{r=1}^{n}ta{n}^{-1}\frac{\left(r+1\right)}{1+r\left(r+1\right)}}}$

$=ta{n}^{-1}2-ta{n}^{-1}+......+ta{n}^{-1}\left(n+1\right)-ta{n}^{-1}n$            

For n  $\infty$ value = $\frac{\pi }{2}-\frac{\pi }{4}=\frac{\pi }{4}$  

$\therefore tan\left(\frac{\pi }{4}\right)=1$           

Number divisible by 3;

(a) sum of the digit must be divisible by 3

(i)  $\boxed{1}\boxed{2}\boxed{3}$  3! = 6

(ii) 1, 3, 5 -> 3! = 6

(iii) 2, 3, 4 -> 3! = 6

(iv) 3, 4, 5 = 3! = 6

Total = 24

(b) Divisible by

$\boxed{}\boxed{}\boxed{5}$  4 * 3 = 12

(c) Now common divisible by both

$\boxed{1}\boxed{3}\boxed{5}$ 2! = 2

For 3, 4  $\boxed{}\boxed{}\boxed{5}$  2! = 2

$\therefore$ Total ways = 24 + 12 – 4 = 32

xyz = 24

24 = 2³ * 3

Let's distribute 2, 3 among 3 variables. No. of positive integral solution =

No. of ways to distribute = 

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

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

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