Permutations and Combinations

Given word is INEONVENIENCE. The distinet letters are I, N, C, O, V, E (6 letters) and their frequencies are
‌I−2
‌N−4
‌C−2
‌O−1
‌V−1
‌E−3
Total number of 5-letters words
(i) When 5 distinct letters, number of words =6P5=720
(ii) One pair and 3 distinct letters,
‌ number of words ‌‌=‌4C1×‌5C3×‌

5!

2!

‌=4×10×60=2400
(iii) One triple and 2 distinct letters, number of words
‌=‌2C1×‌5C2×‌

5!

3!

‌=2×10×20=400

(iv) Two pairs and one distinct, number
‌ of words ‌‌=‌4C2×‌4C1×‌

5!

2!2!

‌=6×4×30=720
(v) One triplet and one pair, number of words
=4×‌

5!

3!2!

=4×10=40
(vi) One 4 a like and one different
=‌1C1×‌5C1×‌

5!

4!

=25
∴ Number of words with at least one repealed letters
‌=(720+2400+400+720+40+25)−720
‌=3585