In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

(A) 736
(B) 768
(C) 792
(D) 840
(E) 876


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IIM CAT Level Question
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three letters can be arranged in 3! ways.
only one combination EIU is required.

7 letters can be arranged in 7! ways.

thus 7!/ 3! * 1 = 840.

D

1. Consider EIU as one. then total possible combination of words = 5* 4! = 120

2. Consider EI and IU together total possible combinations = [5 (positions for EI out of 7) * 5! (rest of the letters)] +

[ 5 (positions for IU)* 5! (rest of the letters)]

Now in half of the combinations U will be ahead of EI and E will be ahead of IU.

So 1/2[5 (positions for EI out of 7) * 5! (rest of the letters)] +1/2 [ 5 (positions for IU)* 5! (rest of the letters)]

total will give = 300 + 300 = 600.

3. Consider E-I-U-- with one letter spacing in between = 3(positions of EIU shifting) * 4! (letters to be arranged) = 72

4. Consider E-I--U- and E---I-U with two letter spacing between IU and EI = 2 * 4! = 48

thus totaling 120 + 600 + 72 + 48 = 840.


Alternately , Consider

all possible letter arrangements = 7!

only one combination out of 3! combinations possible for EIU. meaning 3 are taken in a group,but they are not alike.
Means total combinations remain 7! and not 5! ( EIU together + 4 letters).

So, essentially it is 7P3 =
Hence 7!/ 3! = 840

Darn ! I am not Amt2k9. I think it is pretty obvious that in 1/6th of the ways out of 7! the vowels will be ordered. This reasoning is parallel to the situation when we have 3 people lets say A B C sitting in a row and we want to know the ways in which they will be alphabetical order. There is just one way ie. 3!/6=1 or 3!/3!

The division by 3! is NOT owned by the concept of weeding out the duplicates. We will apply it when we need it

Commenting as I struggled getting my head around the concept. Kudos given above.

There are \(7_P7 = 7!\) ways to arrange the 7 letters.

For each of these combinations, there is a single valid arrangement of the vowels: \(\frac{1}{3_P3}\)

\(7! \times \frac{1}{3!} = \frac{7!}{3!} = \frac{7 \times 6 \times 5 \times 4 \times 3!}{3!} = 840 = D\)
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I have really tried to understand the concepts applied here from all the explanations but I still don't.Can someone be kind enough to help with a link that explains these concepts clearly...:-(

Thanks.

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