sana.noor wrote:How many 3 letter codes can be formed from the PEPPER with replacement.

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

OA is C

There other other counting techniques we could use here, but the relatively small answer 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

satish_iitg wrote:
Dear Brent,

Can you please tell me why 6C3 is wrong.

There are two reasons why 6C3 (combinations) won't work here.

First, since the order of the 3 letters matters (e.g., the word PRE is different from the word REP) we cannot use combinations. We use combinations when the order does not matter (for example, we can select 3 friends from 6 friends in 6C3 ways).

Second, we have duplicated letters (three P's and two E's). To illustrate why duplicated letters messes things up, let's examine what would happen if the question were, "How many 3 letter codes can be formed from PPPPPP?" Here, we can see that 6C3 won't work because all of the letters are the same. In fact, there's only one 3-letter word we can create (PPP). In the same way, the duplicated letters in PEPPER make things trickier.

Cheers,
Brent

sarawutc wrote:I have a question... the question says "with replacement". Doesn't it mean that you can have 3Ps, 3Es, and 3Rs. I'm so confused.

Yeah, I didn't like that phrase, so I just ignored it
I have no idea what the author of the question intended, but you can be certain that an official GMAT question would never contain such ambiguity.

The question, as I read it, is asking asking us to choose 3 letters from PEPPER. So, for example, we can't have three R's.

If the question intended us to be able to use up to 3Ps, 3Es, and 3Rs, then it doesn't make any sense to mention "PEPPER." The question should have just asked, "If duplication is permitted, how many 3-letter codes can be formed using only P's, R's and E's?"

Of course, that still may have been the author's intention. We can't be sure.

Let's answer the question, "If duplication is permitted, how many 3-letter codes can be formed using only P's, R's and E's?"

Take the task of building 3-letter codes and break it into stages.

Stage 1: Select the first letter
We have 3 letters to choose from (P, E, or R), so we can accomplish stage 1 in 3 ways.

Stage 2: Select the second letter
We have 3 letters to choose from (P, E, or R), so we can accomplish stage 2 in 3 ways.

Stage 3: Select the third letter
We have 3 letters to choose from (P, E, or R), so we can accomplish stage 3 in 3 ways.

By the Fundamental Counting Principle (FCP) we can complete all 3 stages (and thus create a 3-letter code) in (3)(3)(3) ways ([spoiler]= 27 ways[/spoiler])

Cheers,
Brent

Aside: For more information about the FCP, we have a free video on the subject: https://www.gmatprepnow.com/module/gmat-counting?id=775

gocoder wrote:

Brent@GMATPrepNow wrote:How many 3 letter codes can be formed from the PEPPER with replacement.

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

OA is C

We can arrange 3 unique objects in 3! ways (6 ways)

Cheers,
Brent

If the Same question had asked to pick 3 letters without replacing, whether the answer would be 6 ?[/quote]

That alteration to the question would really complicate matters, enough to make it out of scope for the GMAT.
The answer would be more than 6 though (using just P's and E's, there are 7 possible arrangements).

Cheers,
Brent