How many 4 letter words can be formed using the letters of the word ‘B

Question

How many 4 letter words can be formed using the letters of the word ‘BALLASTIC’?

A. 1016
B. 1126
C. 1190
D. 1206
E. 1356

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GMATinsight · Answer

Total letters = 9

But, A repeats twice and L repeats Twice

So total Distinct letters = 7

Case 1: All letters are distinct =  ^7C_4*4! = 35*24 = 840

Case 2: Two letters are Same but other two are distinct =  ^6C_2* ^2C_1 * \frac{4!}{2!} = 30*12 = 360

Case 3: Using 2 As and 2 Ls  = \frac{4!}{2!*2!} = 6

Total = 840+360+6 = 1206

Answer: Option D

GreatII · Answer

Shouldnt it be 7^4 ?

sydqur · Answer

I really don't understand this. Please can someone help me?

This is a permutations problem, 9P4 = 3024
There are two letters repeated twice: A and L.

This means that we have to do 3024 / (2! * 2!). Hence the answer is 756.

HarshavardhanR · Answer

- The word BALLASTIC has 2 As, 2 Ls, and 5 other letters (BSTIC).
- We need to find the number of 4 letter words that are possible. 
- There are 3 types of words that can be formed. We need to find the number of possible words in each type and add them up to find the overall number of words.

- Type 1: "PQRS" (all letters distinct).

The word BALLASTIC has 7 distinct letters: B,S,T,I,C,A,L.

-> How many ways can 4 letters be selected from these 4 distinct letters? 7C4
-> Once selected, how many arrangements can be created using those 4 letters? 4!
-> So, overall, how many words with each letter distinct can be created?  7C4 x 4! = 840

- Type 2: "PPQR" (3 distinct letters, one letter repeats once)

-> How many ways to choose between A and L for the letter that will repeat? 2C1
-> Now, we have chosen one among A and L. 
-> How many letters to choose from the available As (or Ls)? 2C2 (we need 2 As or 2Ls from the 2As or 2Ls available)
-> Now, we are left with 6 distinct letters to choose from (B,S,T,I,C and either A or L - whatever was not used from between A and L)
-> How many ways to choose the remaining 2 letters from these 6 distinct letters? 6C2
-> Now, we have our 4 letters out of which one letter repeats.
-> How many arrangements are possible? 4!/2!
-> So, overall, how many words of this type are possible?  2C1 x 2C2 x 6C2 x 4!/2! = 360

- Type 3 "PPQQ" (2 distinct letters, each repeating once)

-> This can only happen when the 2 letters chosen are A and L. 2C2 x 2C2 = 1
-> How many arrangements are possible? 4!/(2! 2!) = 6
-> So, overall, how many words of this type are possible?  1x6 = 6

-> Final answer: 840 + 360 + 6 = 1206. Choice D.

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Harsha
Enthu about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com

How many 4 letter words can be formed using the letters of the word ‘B

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