Permutations and Combinations

Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are$\; \text{I-2}$$\; \text{N-4}$$\; \text{C-2}$$\; \text{O-1}$$\; \text{V-1}$$\; \text{E-3}$Total number of 5-letters words(i) When 5 distinct letters, number of words $=6 P_5=720$(ii) One pair and 3 distinct letters,$\; \text{ number of words } \;\; = {}^4C_1 \times {}^5C_3 \times \frac{5!}{2!}$$\; = 4 \times 10 \times 60 = 2400$(iii) One triple and 2 distinct letters, number of words$\; = {}^2C_1 \times {}^5C_2 \times \frac{5!}{3!}$$\; = 2 \times 10 \times 20 = 400$ (iv) Two pairs and one distinct, number$\; \text{ of words } \;\; = {}^4C_2 \times {}^4C_1 \times \frac{5!}{2!2!}$$\; = 6 \times 4 \times 30 = 720$(v) One triplet and one pair, number of words$=4 \times \frac{5!}{3!2!} = 4 \times 10 = 40$(vi) One 4 a like and one different$= {}^1C_1 \times {}^5C_1 \times \frac{5!}{4!} = 25$$\therefore$ Number of words with at least one repealed letters$\; = (720+2400+400+720+40+25)-720$$\; = 3585$