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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