Permutations and Combinations

EHINTZ

words starting with E is 5!

words starting with H is 5!

words starting with I is 5!

words starting with N is 5!

words starting with T is 5!

$9*9*3=81*3=243$

Single digit factor of $12\to \mathrm{1,2},\mathrm{3,4},6$

  $(x,y,z)$  where , $xyz=12$

$(\mathrm{6,2},1)(\mathrm{4,3},1)(\mathrm{3,2},2)$

$\begin{array}{ccc}\downarrow & \downarrow & \downarrow \\ \text{6Ways}& \text{6 Ways}& \text{3Ways}\end{array}$

 Total 15 Ways

(6!/ (5!1!) * 2! + (6!/ (4!2!) * 2! + (6!/ (3!)²2!) * 2! = 62

All even + 2 odd 1 even
¹? C? + ¹? C? * ¹? C?
(10*9*8)/6 + (10*9)/2 * 10
120 + 450 = 570

$x_1+x_2=6,13\to 6+6=12$

$\begin{array}{l}{x}_1+x_2=4,11,18\to 4+8+1=13\\ x_1+x_2=2,9,16\to 2+9+3=14\\ x_1+x_2=7,14\to 7+5=12\\ x_1+x_2=5,12\to 5+7=12\end{array}$

= 63

I (6)       F (8)

Case I       2            4

Case II      3            6

Case III     4            8

Total =   

All even + 2 odd 1 even
¹? C? + ¹? C? * ¹? C?
120 + 450
=570

There are 7 letters permutation.
The total number of 4 letters
Out of 2A's, 2K's and 3 different letters O, L, T.
= Coefficient of x? in 4! (1 + x + x²/2!)² (1 + x)³
= Coefficient of x? in 4! (1 + x + x²/2)² (1 + x)³
= Coefficient of x? in 4! [ (1 + x)? + (1 + x)? x² + (1 + x)³ + x? /4]
= 4! [? C? +? C? + ³C? /4] = 4! (5 + 6 + 1/4)
= 24 [11 + 1/4]
= 264 + 6 = 270

First we arrange 5 red cubes in a row and assume x? , x? , x? , x? , x? and x? number of blue cubes between them
Here, x? + x? + x? + x? + x? + x? = 11
and x? , x? , x? , x? ≥ 2
so x? + x? + x? + x? + x? + x? = 3
No. of solutions =? C? = 56