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

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Joined: 20 Jul 2017
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How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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03 Feb 2020, 08:41
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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.
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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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03 Feb 2020, 22:49
5
1
Dillesh4096 wrote:
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.

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)
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Director
Joined: 28 Jul 2016
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How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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Updated on: 03 Feb 2020, 20:34
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
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Originally posted by globaldesi on 03 Feb 2020, 10:14.
Last edited by globaldesi on 03 Feb 2020, 20:34, edited 1 time in total.
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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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03 Feb 2020, 20:15
globaldesi wrote:
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
=$$\frac{ 7*6 (5!)}{2!}- \frac{5!*3!}{2!}$$
= 1560
hence none of these

globaldesi

You have done the calculation mistake.
=$$\frac{ 7*6 (5!)}{2!}- \frac{5!*3!}{2!}$$
= 2520-360
= 2160
IMO-B
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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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03 Feb 2020, 20:31
rajatchopra1994 wrote:
globaldesi wrote:
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
=$$\frac{ 7*6 (5!)}{2!}- \frac{5!*3!}{2!}$$
= 1560
hence none of these

globaldesi

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

Ahh ohh.
Thanks for correcting.
I am bad at calculations.
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How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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03 Feb 2020, 21:58
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
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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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05 Feb 2020, 10:06
yatharthdas wrote:
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

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 )
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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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09 Feb 2020, 05:17
Dillesh4096 wrote:
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.

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![((7 x 6)/2) - 3]

5!(21 - 3)

5! x 18 = 2160

Answer: B

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Re: How many 7 alphabetic words can be formed using all the alphabets of  [#permalink]

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09 Feb 2020, 06:59
An interesting mathematical problem. I have always understood the exact sciences more than literature and languages. So with writing essays, I was always helped
Re: How many 7 alphabetic words can be formed using all the alphabets of   [#permalink] 09 Feb 2020, 06:59
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# How many 7 alphabetic words can be formed using all the alphabets of

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