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Combinations Problem

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Combinations Problem

by topspin360 » Sat Dec 22, 2012 10:10 am
Problem: Possible arrangements for the word REVIEW if one E can't be next to the other.

Answer is:

6!/2!. I understand how this is done.

I thought 6! - 5! should also work but don't understand why it doesn't. 'Total combinations' - 'Combinations where two Es are together' would give us 6! - 5!. Can anyone explain what's the flaw in my reasoning?

Thanks.
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by The Iceman » Mon Dec 24, 2012 4:11 am
topspin360 wrote:Problem: Possible arrangements for the word REVIEW if one E can't be next to the other.

Answer is:

6!/2!. I understand how this is done.

I thought 6! - 5! should also work but don't understand why it doesn't. 'Total combinations' - 'Combinations where two Es are together' would give us 6! - 5!. Can anyone explain what's the flaw in my reasoning?

Thanks.
To your surprise none of these answers would be correct.

Solution I: 6!/2 (total arrangement of letters of review) - 5!(if E were placed together) = 240

Solution II: There are 5 slots correspoding to other 4 letters. We need to chose 2 slots out of these 5 in which we can fix the Es. This is done in 5C2 ways. The remaining 4 letters can arrange themselves in 4! ways.

So total number of arrrangements: 5C2 * 4! = 10*24 = 240
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by tabsang » Mon Dec 24, 2012 6:42 am
topspin360 wrote:Problem: Possible arrangements for the word REVIEW if one E can't be next to the other.

Answer is:

6!/2!. I understand how this is done.

I thought 6! - 5! should also work but don't understand why it doesn't. 'Total combinations' - 'Combinations where two Es are together' would give us 6! - 5!. Can anyone explain what's the flaw in my reasoning?

Thanks.
I use the following approach for the "not together" type of questions. Always works :)
Condition: The 2 E's cannot be placed next to each other.
Arrange the remaining letters in the following way:
_R_V_I_W_
R,V,I & W can be arranged in 4!=24 ways.
There are 5 slots for the 2 E's and they can be arranged in 5C2 = 10 ways
Total combinations = 24x10=240 ways

P.S.: if instead of the 2 E's, we had two distinct items, we would've multiplied by 2! to account for the number of ways those two distinct items can be arranged.

Hope this Helps.

Cheers,
Taz
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by The Iceman » Mon Dec 24, 2012 8:22 am
tabsang wrote:P.S.: if instead of the 2 E's, we had two distinct items, we would've multiplied by 2! to account for the number of ways those two distinct items can be arranged.
You bet :)
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