Combinations Problem

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topspin360: Master | Next Rank: 500 Posts; Posts: 103; Joined: Sat Jun 02, 2012 9:46 pm; Thanked: 1 times

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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Source: Beat The GMAT — Problem Solving | Sat Dec 22, 2012 10:10 am

The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members; Location: India; Thanked: 47 times

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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tabsang: Senior | Next Rank: 100 Posts; Posts: 64; Joined: Mon Jun 13, 2011 5:27 am; Thanked: 5 times

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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The Iceman: Master | Next Rank: 500 Posts; Posts: 194; Joined: Mon Oct 15, 2012 7:14 pmFollowed by: 6 members; Location: India; Thanked: 47 times

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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