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

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