karan.7045 wrote:How to solve the permutations and combination questions such as

How many 3 letter codes can be formed from the PEPPER with replacement.

a)6
b)18
c)19
d)20
e)36

The relatively small answers choices suggest that listing/counting might be the fastest approach.
Let's examine the different cases.

All 3 letters the same
Only possible with three P's
PPP (1 code)


2 letters the same, 1 different
- 2 P's and 1 R
RPP, PRP, PPR
(3 codes)

- 2 P's and 1 E
(3 codes)

- 2 E's and 1 P
(3 codes)

- 2 E's and 1 R
(3 codes)

All 3 letters different (1 P, 1 R and 1 E)
We can arrange 3 unique objects in 3! ways (6 ways)
(6 codes)

Add them up: 1+3+3+3+3+6 = [spoiler]19 = C[/spoiler]

Cheers,
Brent