How many 7 alphabetic words can be formed using all the alphabets of

Question

How many 7-letter words can be formed using all the alphabets of the word  SIMILAR  given the condition that all the vowels are not together.

A. 1800
B. 2160
C. 4320
D. 4680
E. None of these.

KarishmaB · Accepted Answer

7 letter words using all 7 letters of SIMILAR = 7!/2! (because I repeats)

7 letter words using all 7 letters such that all vowels are together:
Tie the vowels together into 1 - IIA which can be written in 3 different ways.
Now we need to arrange 5 different letters/groups - S, M, L, R, IIA which can be done in 5! ways.
So total such words = 5! * 3

Words in which vowels are not together = 7!/2! - 5!*3 = 18*5! = 2160

Answer (B)

globaldesi · Answer

total different numbers formed are:
 7!/2! 
divide by 2! since I repeats twice
assume all vowels are together
hence final total words will be 
 5!*3!/2! 
hence all vowels no together 
total case - all together
= \frac{ 7*6 (5!)}{2!}- \frac{5!*3!}{2!} 
= 2160
hence B

rajatchopra1994 · Answer

globaldesi

You have done the calculation mistake.
= \frac{ 7*6 (5!)}{2!}- \frac{5!*3!}{2!} 
= 2520-360
= 2160
IMO-B

globaldesi · Answer

Ahh ohh.
Thanks for correcting.
I am bad at calculations.

yatharthdas · Answer

I THINK IT SHOULD BE LIKE THIS....
total no. of arrangements possible = 7p7 or 7!=5040
and total no. of arrangements with vowels together = (all wovels together as a pack)*(vowels as one and other letters)=3!*5!=720

=5040-720=(4320) Ans.

M not sure pl. Help Experts

globaldesi · Answer

what about same words such as II
ITI is same as ITI .. 
IIT and IIT are again 
from IIT you can only form 3 words
IIT , ITI, TII
that 3!/2! (total words divided by number of similar words )

ScottTargetTestPrep · Answer

If the letters were all different, the total number of ways to arrange the letters would be 7! . However, since there are 2 identical I’s, we divide by 2!, so we have 7!/2! = 7!/2 ways.

If all three vowels are together, we can treat them as a single item and arrange that group of vowels with the remaining 4 consonants, obtaining 5! ways. In addition, the three vowels I - I - A can be arranged in 3!/2! = 3 ways. Thus, the total number of arrangements in which all the vowels are together is 5! x 3.

Thus, the number of 7-letter words that can be formed such that the vowels are NOT all together is the difference:

7!/2 - 5! x 3

We see that 5! is a common factor of each term, so we have:

5!

5!(21 - 3)

5! x 18 = 2160

Answer: B

fiona243 · Answer

An interesting mathematical problem. I have always understood the exact sciences more than literature and languages. So with writing essays, I was always helped

CrackverbalGMAT · Answer

The number of 7 letter words = 7!/2! (2 I's)

=2520

The 3 vowels when grouped are IIA and if we consider them as one alphabet,then we need to arrange S,M,L,R and IIA.

These 5 alphabets(considering IIA as an alphabet) can be arranged in 5! * 3!/2! (IIA arranges in 3!/2!)

Thus total arrangements = 120 * 3=360

Number of arrangements where vowels are not together = 2520 - 360

= 2160 
 (option b)

Devmitra Sen
GMAT SME

gmatflunkie · Answer

KarishmaB 
 CrackverbalGMAT

My approach was different and off by a factor of 3.

I tried placing the consonances in 4 spots in 4! ways.

Then, place the vowels in the 5 gaps. 5C3 ways

Then, arrange the vowels. 3!/2!

|C|C|C|C|

24*10*3 = 720

Why is this wrong?

KarishmaB · Answer

The question gives the condition 'all vowels are not together' not 'there must be atleast one consonant between any two vowels'
So we put all the vowels together and remove this case from total. In the total SIMIALR is a an acceptable word. 
It is not acceptable the way you have calculated.

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How many 7 alphabetic words can be formed using all the alphabets of

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