Last visit was: 14 Jul 2024, 06:00 It is currently 14 Jul 2024, 06:00
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# How many distinct 4-letter “words” (with or without meaning) can be ma

SORT BY:
Tags:
Show Tags
Hide Tags
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3139
Own Kudos [?]: 5008 [9]
Given Kudos: 1859
Location: India
Manager
Joined: 25 Sep 2019
Posts: 56
Own Kudos [?]: 31 [2]
Given Kudos: 103
Intern
Joined: 10 Dec 2022
Posts: 1
Own Kudos [?]: 1 [0]
Given Kudos: 108
Location: Switzerland
Quant Chat Moderator
Joined: 22 Dec 2016
Posts: 3139
Own Kudos [?]: 5008 [2]
Given Kudos: 1859
Location: India
How many distinct 4-letter words (with or without meaning) can be ma [#permalink]
2
Kudos
olaflesniak1998 wrote:
Could you maybe elaborate why you in case 2 and 3 multiply by factorial 4! ?

We have the following combinations of alphabets available with us -

E X M T O A A I I N N

Question: Create four-letter words with or without meaning.

The creation of a letter involves the following two steps -

Step 1: Select four letters

Step 2: Arrange those four letters

Case 1: No letters repeat (4 distinct alphabets)

Step 1: Select four letters

There are 8 distinct letters, hence we can select 4 letters in $$^8C_4$$ ways.

Step 2: Arrange those four letters

The 4 distinct letters so selected in step 1, can be arranged in 4! ways.

Possible words : $$^8C_4$$ * 4! = 1680

Case 2: One pair of alphabet repeats (2 distinct alphabets & 1 repeating pair)

Step 1: Select four letters

The repeating pair can be selected in $$^3C_1$$ ways

From the remaining 7 distinct alphabets, we can select two alphabets in $$^7C_2$$ ways

Step 2: Arrange those four letters

The word is of the form XXYZ, and the number of possible arrangements = $$\frac{4!}{2!}$$

XX is the repeating pair
Y & Z are unique alphabets

Possible words : $$^3C_1 * ^7C_2 * \frac{4!}{2!}$$ = 756

Case 3: Two pairs of alphabet repeats (2 pairs of repeating alphabets)

Step 1: Select four letters

The repeating pairs can be selected in $$^3C_2$$ ways

Step 2: Arrange those four letters

The word is of the form XXYY, and the number of possible arrangements = $$\frac{4!}{2!*2!}$$

XX & YY are the repeating pairs

Possible words : $$^3C_2 * \frac{4!}{2!*2!}$$ = 18

Total of all three cases = 1680 + 756 + 18 = 2454

Option B