In how many different ways, can the letters of the word 'INHALE' be arranged?

ⁿP₃ = 9240 ? n! / (n-3)! = 9240
? n(n - 1)(n - 2) = 9240
? n(n - 1)(n - 2) = 22 x 21 x 20
? n = 22

By formula, ⁿc_x = ⁿc_y
? x = y or x + y = n

Now, ⁵⁰c_r = ⁵⁰c_{r + 2}
? r + r + 2 = 50 or r = r + 2
? 2r = 48 [? r = r + 2 is not possible]
? r = 24

⁵ P₂ = 5! / (5 - 2)! = 5 x 4 = 20

ⁿ⁺²C₈ : ^n-2P₄ = 57 : 16
? {(n + 2 )! (n - 6)!} / {(n - 6)! (n - 2)! 8!} = 57/16
? (n + 2) (n + 1) n(n - 1) = 143640
? (n² + n - 2) (n² + n ) = 143640
? (n² + n )² - 2(n² + n) + 1 = 143641
? (n² + n - 1)² = (379)²
[? n² + n - 1 > 0]
? n² + n - 1 = 379
? n² + n - 380 = 0
? (n + 20) (n - 19) = 0
?n = 19 ( Since n is not negative)

Required no. of hand shakes = {8 x (8 - 1)}/2 = 28

Number of arrangements n ! /( p ! q ! r ! )
Total letters = 6, but R has come twice.
So, required number of arrangements
= 6 ! / 2 ! = (6 x 5 x 4 x 3 x 2 !) / 2 ! = 360

Required number of arrangements = 6 ! / 3 ! [? S has come thrice ]
= [ 6 x 5 x 4 x 3! ] /3 !
= 120

Total letters = 7, but N has come twice.
So, required number of arrangements = 7 ! / 2 !
= [7 x 6 x 5 x 4 x 3 x 2 ! ] / 2 ! = 2520

Total number of letters = 7, but N has come twice.
So, required number of arrangements = 7 ! / 2 !
= [7 x 6 x 5 x 4 x 3 x 2! ] / 2!
= 2520

The required different ways = 7 ! / 2 !
= 7 x 6 x 5 x 4 x 3 x 2!/2!
= 2520