Martha and Four Friends
Date: 09/17/2001 at 07:30:12 From: vidya Subject: Combinations permutations Martha and four friends go to a movie. How many different ways can they sit together with Martha always between two friends? My answer is 72, whereas the solutions gives the answer to the question as 24. Thanking you in anticipation, Vidya
Date: 09/17/2001 at 09:56:06 From: Doctor Ian Subject: Re: Combinations permutations Hi Vidya, I get the same answer you do. The total number of ways that they can sit together is 5! = 5 * 4 * 3 * 2 * 1 = 120 The total number of ways that Martha can sit on the left end is 4! = 24 The total number of ways that she can sit on the right end is 4! = 24 So the total number of ways that she can sit in the middle must be 120 - 24 - 24 = 72 I'm guessing that the person who said that 24 is the answer reasoned this way: There are 3*2*1 = 6 ways that Martha can sit with two of her friends on the inside. Then there are 4 ways to choose the person on the left; which leaves just 1 way to choose the person on the right; for a total of 4*3*2*1*1 = 24 ways for the friends to sit together. However, this is incorrect, since it doesn't take into account the ways that the inner and outer groups can be chosen. But as I said, I'm just guessing. But in problems like this - assuming they are small enough - there is always a way to settle a question definitively, which is to enumerate all the possibilities. To generate them systematically, I'm going to use the notation '-- ---', in which the left side represents choices that have been made, and the right side represents the unchosen elements. And let's say that Martha is 'a', and mark each permutation in which she doesn't appear on either end. a bcde -> ab cde -> abc de -> abcde abced abd ce -> abdce abdec abe cd -> abecd abedc ac bde -> acb de -> acbde acbed acd be -> acdbe acdeb ace bd -> acebd acedb ad bce -> adb ce -> adbce adbec adc be -> adcbe adceb ade bc -> adebc adecb ae bcd -> aeb cd -> aebcd aebdc aec bd -> aecbd aecdb aed bc -> aedbc aedcb b acde -> ba cde -> bac de -> bacde * baced * bad ce -> badce * badec * bae cd -> baecd * baedc * bc ade -> bca de -> bcade * bcaed * bcd ae -> bcdae * bcdea bce ad -> bcead * bceda bd ace -> bda ce -> bdace * bdaec * bdc ae -> bdcae * bdcea bde ac -> bdeac * bdeca be acd -> bea cd -> beacd * beadc * bec ad -> becad * becda bed ac -> bedac * bedca c abde -> ca bde -> cab de -> cabde * cabed * cad be -> cadbe * cadeb * cae bd -> caebd * caedb * cb ade -> cba de -> cbade * cbaed * cbd ae -> cbdae * cbdea cbe ad -> cbead * cbeda cd abe -> cda be -> cdabe * cdaeb * cdb ae -> cdbae * cdbea cde ab -> cdeab * cdeba ce abd -> cea bd -> ceabd * ceadb * ceb ad -> cebad * cebda ced ab -> cedab * cedba d abce -> da bce -> dab ce -> dabce * dabec * dac be -> dacbe * daceb * dae bc -> daebc * daecb * db ace -> dba ce -> dbace * dbaec * dbc ae -> dbcae * dbcea dbe ac -> dbeac * dbeca dc abe -> dca be -> dcabe * dcaeb * dcb ae -> dcbae * dcbea dce ab -> dceab * decba de abc -> dea bc -> deabc * deacb * deb ac -> debac * debca dec ab -> decab * decba e abcd -> ea bcd -> eab cd -> eabcd * eabdc * eac bd -> eacbd * eacdb * ead bc -> eadbc * eadcb * eb acd -> eba cd -> ebacd * ebadc * ebc ad -> ebcad * ebcda ebd ac -> ebdac * ebdca ec abd -> eca bd -> ecabd * ecadb * ecb ad -> ecbad * ecbda ecd ab -> ecdab * ecdba ed abc -> eda bc -> edabc * edacb * edb ac -> edbac * edbca edc ab -> edcab * edcba I count 72 different ways for them to sit together with Martha in the middle. Does this help? - Doctor Ian, The Math Forum http://mathforum.org/dr.math/
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