How many 3-letter codes can be formed by choosing, without replacement

Question

How many 3-letter codes can be formed by choosing, without replacement, 3 letters from the word PEPPER? A) 6 B) 18 C) 19 D) 27 E) 30 PEPPER.png

stne · Accepted Answer

Excellent solution already posted above, just putting the same in a different way below:

Only these cases are possible:

PER  = 3! =6

EER  = \frac{3!}{2!} = 3

EEP  = \frac{3!}{2!} = 3

PPE  = \frac{3!}{2!} = 3

PPR  = \frac{3!}{2!} = 3

PPP  = 1

Total  = 6+ 3+3+3+3+1 = 19

Ans C

Hope it helps.

gmatophobia · Answer

P E R
P E
P

We see that the letters repeat. Hence, we have to take various cases to solve this question. In each case, we will perform two steps

Step 1: Selecting the letters
Step 2: Arranging the letters

Case 1  : No Alphabet Repeats

Step 1: Selecting the letters

We have three distinct letters, and we need to select all of them. They can be selected in  ^3C_3  = 1 way.

Step 2: Arranging the letters

As the letters are distinct, the number of ways they can be arranged = 3! = 6 ways

Sub Total : 6 * 1 = 6 ways

Case 2  : One Alphabet repeats twice

Step 1: Selecting the letters

The repeating alphabet can be either 'P' or 'E'. We need to choose one of them. The selection can be made in  ^2C_1  ways.

Once the selection has been made, out of the remaining two alphabets, we have to choose one.

The third alphabet can be chosen in again  ^2C_1  ways.

Step 2: Arranging the letters

The letter is in the form of XXY. Hence the number of ways we can arrange =  \frac{3!}{2!} = 3  ways

Sub-Total = 2 * 2 * 3 = 12 ways

Case 3  : One Alphabet repeats thrice

Step 1: Selecting the letters

The repeating alphabet can be only 'P'. The selection can be made in  ^1C_1  way.

Step 2: Arranging the letters

The only possible arrangement is PPP. Hence, the letters can be arranged only in 1 way.

Sub-Total = 1

Total = Case 1 + Case 2 + Case 3 = 6 + 12 + 1 = 19

Option C

KarishmaB · Answer

A 3 letter code out of PEPPER can be made in 3 different ways. All 3 letters the same. The code is PPP. Nothing else is possible. So 1 code. 2 letters same. Select the letter to be repeated in 2 ways (either P or E). Select the third letter in 2 ways (R or leftover letter). Arrange all letters in 3 different ways (e.g. PPR, PRP, RPP) Total codes = 2 * 2 * 3 = 12 No letter same. We have only 3 distinct letters (P, E, R) so all 3 must be taken. We can arrange them in 3! ways = 6 codes Total number of codes = 1 + 12 + 6 = 19 Answer (C) Video on Permutations: https://youtu.be/LFnLKx06EMU Video on Combinations: https://youtu.be/tUPJhcUxllQ

DanTheGMATMan · Answer

The trick is to see that you have to differentiate scenarios based on whether you repeat identical letters and how many:

https://www.youtube.com/watch?v=GrPi3jYtyAQ

sahil770 · Answer

I generated this response with AI. I founded it intutive so sharing here.

Imagine a game where you have three slots to fill with letters from the magical word PEPPER. But beware, once a letter is plucked, it vanishes from the word!

The Case of the Triplets

First, consider the scenario where all three slots are filled with the same letter. Only the letter 'P' exists in triplicate, so there's just 1 way to achieve this (PPP).

The Twin's Tale

Next, envision two slots holding identical twins, while the third slot houses a solitary letter. The twins can be either 'P' or 'E'. Once the twins are chosen, only two options remain for the lone wolf letter. And these three letters can be arranged in 3 ways (accounting for the identical twins). This gives us 2 choices for twins * 2 choices for the loner * 3 arrangements = 12 possibilities.

The Trio of Uniqueness

Finally, picture each slot holding a unique letter. The only such trio is 'PER'. These three distinct letters can be shuffled in 6 ways (3!).

The Grand Summation

To find the total number of codes, we gather the possibilities from all these scenarios: 1 (triplets) + 12 (twins) + 6 (unique trio) = 19

Thus, there are 19 mystical codes that can be formed from the enchanted word PEPPER!

QuantMadeEasy · Answer

This video solution could be helpful

https://www.youtube.com/watch?v=71L8XnTjH7w

Smart Prep (Tutor)

Ophelia__ · Answer

Bunuel  can you suggest any similar questions to this one, please? Thank you so much

Bunuel · Answer

Check  Constructing Numbers, Codes and Passwords Questions  in our  Special Questions Directory .

Hope it helps.

LeelooMats · Answer

Hello, can you please tell me why the repeating alphabet in 2nd case cannot be "R". Thank You.

Bunuel · Answer

Because there is only one "R" in PEPPER, we are choosing three letters from PEPPER without replacement, so we cannot choose two "R"s.

einstein801 · Answer

Hi  KarishmaB , i dont really follow - is there no Choosing here? e.g. for ERP: 2C1*1C1*3C1 and finally *3! for the different arrangements

kanwar08 · Answer

KarishmaB   Bunuel

I was able to arrive at the answer but in between, I got confused because of the following approach:

I tried to use the fundamental counting principle to arrive at the scenarios. For example for the scenario:

PPE
or 
PPR

We have three spaces to fill from 6 alphabets with 3 unique ones being P, E and R.

While filling the first space, I used simple FCP that the first can be filled in 3 ways (out of 3 Ps) and 2nd space in 2 ways (out of the remaining 2) and the last in 2 ways (E or R).

3*2*2 = 12

Now for their arrangements, we can multiply 12 by 3!/2! i.e. 3
i.e. total possible codes being 36.

Can you please help me understand why FCP is not working over here? Is it all 3 P's the same? Please clarify.

KarishmaB · Answer

FCP is not working because your subsequent selection depends on what you chose previously. Say you selected the first letter in 3 ways. (P, E or R).
Now in how many ways can you select the next one? Well, if you selected R for the first position, you have 2 options (P, E). 
But if you selected P for the first position, you still have 3 options (P, E, R). Hence FCP fails. That is why we need to split the possibilities into 3 distinct groups as done.

kanwar08 · Answer

Hi  KarishmaB ,

I get your point that why FCP can't be used on the overall problem but can we still use it on the subset scenarios.

Eg. One sub-scenario we have is  PPE.

_ , _ , _

The first space can be filled in 3 ways (out of 3 P), second space in 2 ways (out of remaining 2) and the last E in 1 way.

i.e. 3*2*1 = 6
But, do we need to further multiple it by 3!/2! to accommodate different arrangements or FCP covered all those?

Now If I use selection and arrangement separately for this arrangement, my answer comes to be different:

Selection: 3c1 * 2c1*2c1 
Arrangement: 3!/2!
Total = 3*2*2 *3=36
(Can't understand why it's not working)

Same for  PER

_ , _, _

Using FCP, the First space in 3 ways, the second in 2 way, and the third in 1 way i.e. again total 6 ways

3*2*1 = 6

Moreover, if I go with the explanation given by GMATphobia, which uses selection and arrangement separately. Why did he consider unique elements (P,E,R) instead of base (P,E,P,P,E,R) to arrive at

Case 1   : No Alphabet Repeats

Step 1: Selecting the letters

We have three distinct letters, and we need to select all of them. They can be selected in  
  3C3  
 = 1 way.

Why not 3C1 (P) & 2C1 (E) and 1C1(Q). Since we can select any P out of given 3

KarishmaB · Answer

Here is the error you are making - We have 3 Ps but they are identical. I can select a P in 1 way only, not 3C1. They are all P, P, P
How do I know which P I have selected? When you have 3 identical marbles, there is only 1 way in which you can pick 1. There is only 1 way in which you can pick 2 etc.

GmatKnightTutor · Answer

From PEPPER, we can list all the different 3-letter GROUPS that are possible (for example, 2 Ps and 1 E).

PPP
PPR
PPE
PER
ERE
EPE

Remember, these are groups. We therefore have to consider how many different ways the letters in each of these groups can also be arranged. This is because PPR, for example, can be arranged as PRP and be considered a different code.

PPP -> 1 
PPR - > 3! / 2! = 3
PPE -> 3! / 2! = 3
PER -> 3! = 6
ERE -> 3! / 2! = 3
EPE -> 3! / 2! = 3

1+ 3 + 3 + 3 + 6 + 3 + 3 = 19 possible codes

(C) is your answer.

nikita.nanda1810 · Answer

Hi  Bunuel   KarishmaB

I had a question regarding the language of 'without replacement'. I understand how to find the number of 3-letter codes from the letters of the word 'PEPPER' but during my mock I was distracted by the words 'without replacement' - since this is something we usually use for probability. Does this hold any specific significance with respect to this question and if this appears in the context of P&C, how should we view and process this?

Thanks!

KarishmaB · Answer

The words "without replacement" are used to give extra clarity. They mean to say that once you use a letter from these : PEPPER, you cannot use it again i.e. you have only 3 Ps, 2 Es and 1 R. You do not have 3 Es or 3 Rs. You cannot replace them (put them back after using) and pick them again. So think of them as individual tiles with letters on them. Once you use a tile, you do not replace it with another. Now you have only 5 tiles left. 
It means that as you use items from the given list, they deplete. If to start with you have 6 items in the list, as you use they come down to 5 then 4 etc.
"With replacement" means that once you use an item, it is replaced by another same item - or in other words, your list never depletes. If to start with you have 6 items in the list, as you use, the list still has 6 items only.

Alilee · Answer

First, notice the Q is testing on: 
 
 Repeated letters  
 Permuation

This means there are case scenarios that need to be considered: 
 
   no repeated letters    
 
 3*2*1=  6    
 
   1 repeated letters   
 
 P is repeated once & R or left over letter 
 
 PPE
 
 # of way to arrange this is 3*1*1= 3   
 PPE, PEP, EPP  
 
 PPR
 
 # of way to arrange this is 3*1*1= 3   
 PPR, PRP, RPP

E is repeated once & R or left over letter 
 
 EEP 
 
 # of way to arrange this is 3*1*1= 3   
 EEP, EPE, PEE  
 
 EER
 
 # of way to arrange this is 3*1*1= 3   
 EER, ERE, REE

all repeated letters  
 
 PPP <- only  1  option

6+3+3+3+3+1=  19

Whenever there are repeated letters and order matters -> think of it as cases of different scenarios that need to be considered.

How many 3-letter codes can be formed by choosing, without replacement

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How many 3-letter codes can be formed by choosing, without replacement

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