How many 3 letter words (meaningful or meaningless) can be formed usin

Question

How many 3 letter words (meaningful or meaningless) can be formed using letters from the word INTELLECTUAL?

A) 336
B) 370
C) 399
D) 400
E) None of these

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CrackverbalGMAT · Answer

INTELLECTUAL contains 3 L's, 2 E's, 2 T's and 1 each of I, N, C, U and A

There are 8 distinct letters

Number of words having all three letters the same  = 1 (LLL)

Number of words having 2 similar and 1 different

Choose 1 letter out of 3 letters (E, L or T) for 2 similar letters = 3C1 = 3
Choose 1 letter out of the remaining 7 letters for the different letter = 7C1 = 7
Arrange these 3 letters chosen letters in 3!/2! ways = 3 ways (we divide by 2!, as 2 of the letters are the same)

Number of words = 3 * 7 * 3 = 63 words

Number of words where all the letters are different.

Choosing 3 letters out of 8 distinct letters and arranging them = 8C3 * 3! = 336

Total Number of words = 1 + 63 + 336 = 400

Option D

Arun Kumar

GMATinsight · Answer

Frequency of letters
I - 1
N - 1
T - 2
E - 2
L - 3
C - 1
U - 1
A - 1

3 letter words

Case 1 : All letters identical (LLL)
 1 ways

Case 2 : Two letters identical and third different (LLA)
3*7*3 = 63 ways

Case 3 : All letters distinct (INT)
8C3*3! = 56*6 = 336 ways

Total =  1 + 63 + 336  = 400

Answer: Option D

Archit3110 · Answer

INTELLECTUAL
total 12 letters where repetition of T is 2 ; E is 2 ; and L is 3 and rest all 5 letters are 1 ; 
condition given is 3 letter words 
case 1; 
all are similar letters ; only possibility is LLL ; 1
case 2 ; 
two letters are similar and 1 is different ; we can have TTX ; EEX ; LLX ; 
ways to arrange 3!/2! ; 3 
ways to select one of from TT, EE, LL; 3c2 ; 3
and that would leave us with option to chose 1 other letter from total 8 distinct letters; 7c1 
we have 3*3*7 ; 63 ways
case 3
all are different 
total distinct letters 8 ways to chose ; 8c3 ; 56 and this can be arranged in 3! ways ; 56*6 ; 336
total possibilities ;

336+63+1; 400
OPTION D

funkyakki · Answer

hi, i would like to know what's wrong with choosing 3 letters out of 12 letters i.e 12C3 = 220.

Can anybody please explain where i am going wrong ?

GMATinsight · Answer

Two mistakes
1) We need to take arrangements of chosen letters as well which is not taken care of by 12C3
2) The arrangement in case where all letters are different is 3! But in other case where two letters of three are same will have 3!/2! Arrangements hence you can not club these three cases as explained in my explanation.

I hope this help!  funkyakki

samarpan.g28 · Answer

I have grouped the letters like I N (TT) (EE) (LLL) C U A.
There will be three cases to get the three-letter word.
Case 1: When all the three letters are different, we can choose from I N T E L C U A i.e. 8C3*3!=336.
Case 2: When two letters are same, we can choose from (TT) (EE) (LLL) i.e. 3C2 and choose the other letter from the remaining of (TT) (EE) (LLL) I N C U A i.e. 7C1. Then arrange in 3!/2 since two letters are same so 3C2*7C1*3!/2=63 ways.
Case3: When all letters are same, this can be chosen in 1 way only (LLL).
Total 336+63+1=400. Option (D) is correct.

HarshavardhanR · Answer

The word "INTELLECTUAL" contains:

12 letters in total (i, n, 2t, 2e, 3l, c, u, a) 
 8 unique letters (i, n, t, e, l, c, u, a)

We need to find the number of 3-letter words that can be formed.

Case 1: all 3 letters are the same

There is only 1 letter which exists 3 times. LLL - is the only word possible here.

Therefore, 
Number of such words = 1 word

Case 2: 2 letters are the same, 1 letter is different

Which letter will repeat? There are only 3 possible letters that could repeat (t, e, and l). Hence, from 3 possible letters, we need to select that one letter that will appear twice ---- 3C1 
 Which letter will not repeat? Now, from the 8 unique letters available, whichever letter was picked in the above step cannot be again picked. Hence, the letter from above is out. That leaves us with 7 unique letters, from which we need to choose one --- 7C1 
 Now, we have chosen our letters ({t,t,e} for example). How many different words are possible using these letters? This is nothing but number of possible arrangements --- 3!/2!

Therefore,
Number of such words = 3C1 x 7C1 x 3!/2! = 3 x 7 x 3 = 63 words.

Case 3: All 3 letters are different

From 8 unique letters, we need to select 3 different letters --- 8C3 
 Now, we have 3 unique letters ({t,c,e} for example). How many different words are possible using these letters? This is nothing but number of possible arrangements --- 3!

Therefore,
Number of such words = 8C3 x 3! = 336 words

Total number of words = 1 + 63 + 336 =  400 words . Choice D.

---
Harsha

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