Example 4. In how many distinguishable permutations are possible with the letters of the
word PALAKPAKIN?
Solution:
Since the word "distinguishable" is already mentioned in the problem,
obviously the formula that you are going to use is:
n!
P =
p! ! !
There are 10 letters in the word. 2 Ps are alike, 3 A's are alike, 2 K's are alike,
therefore, we have :
10!
(10)(9)(8) 7)(6)(5)45(212)(1)
= (10)(9)(8)(7)(
65) = 151. 200
2! 3! 2!
(25(1)(3)(21(2)(1)
P =
151, 200 ways​