How many words (with or without meaning) can be formed using all the

The question is asking the total number of arrangements possible with the letters of the word “SELFIE” where two E’s are not together.

Arrangements when two E’s are not together = Total arrangements - Arrangements when two E’s are together


In total there are 6 letters but two are identical. so we can arrange in 6! ways. but we divide for those objects that are identical. so divide by 2!. Hence,
Total arrangements = 6!/2!

Now two E's are coupled together. Consider this couple (EE) as one letter. apart from this there are 4 more letters. so we can arrange these 5 different objects in 5! ways.

Two E's can arrange themselves in 2! ways, but we divide for those objects that are identical. so divide by 2!. so arrangement for E's would be 2!/2!.
Hence, Arrangements when two E’s are together = 5! * (2!/2!)


Arrangements when two E’s are not together = 6!/2! - 5! = 5! * ( 6/2 -1 ) = 120 * 2 = 240.

Option E is correct!

Please refer to the picture for the solution.






Saunak Dey

JAMBOBREE OFFICIAL SOLUTION:

We do not have a direct formula to calculate the answer , but we can say

Required answer = total number of words which can be formed – number of words when the two E”s are together

Total number of words = 6! / 2! = 360

To calculate the number of ways in which two e’s are together we will group E’s together .we will get SLFI(EE)
number of ways we can arrange SLFI(EE) = 5!

Number of words when two E’s are together = 5!

Required answer = 360 – 120
= 240

Sorry... i dont understand why we need to divide the 6! by 2! .... Is it because we have 2 E? If we have 3 E, we will divide it by 3! ?

Yes, for SELFIEE the number of arrangements would be 7!/3! because there are 7 letters out of which 3 are identical.


THEORY:

Permutations of \(n\) things of which \(P_1\) are alike of one kind, \(P_2\) are alike of second kind, \(P_3\) are alike of third kind ... \(P_r\) are alike of \(r_{th}\) kind such that: \(P_1+P_2+P_3+..+P_r=n\) is:

\(\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}\).

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is \(\frac{6!}{2!2!}\), as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be \(\frac{9!}{4!3!2!}\).

21. Combinatorics/Counting Methods

• Theory
  Math: Combinatorics
  Combinatorics Made Easy!
  1 hour Combinatorics Video Lesson
  Combinations while dealing with similar and dissimilar things
  When Permutations & Combinations and Data Sufficiency Come Together
  
  • Questions
    Advanced Topic: Probability and Combinations Questions With Solutions
    Letter Arrangements in Words
    Seating Arrangements in a Row and around a Table
    Constructing Numbers, Codes and Passwords
    DS Questions
    PS Questions

For more:
ALL YOU NEED FOR QUANT ! ! !
Ultimate GMAT Quantitative Megathread


Hope it helps.

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