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in statistics list all the ordered arrangements of 5 objects a, b, c, d, and e

Statistics
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in statistcs list all the ordered arrangements of 5 objects a, b, c, d, and e

Apr 6th, 2015

This is a LOT to list. There will be 5! = 5 x 4 x 3 x 2 x 1 = 120 arrangements. To see why:

Start with {abcde}

(2 options for the last two elements, here de or ed. Here's the second):

{abced}

(3 options for the middle element, with each giving 2 options for the last two elements, like above. So here's the remaining 4):

{abdce}

{abdec}

{abecd}

{abedc}

(4 options for the second element, with each giving the 6 combinations we've got above from rearranging the final 3 elements. Here's the remaining 18):

{acbde} {acbed} {acdbe} {acdeb} {acebd} {acedb}

{adbce} {adbec} {adcbe} {adceb} {adebc} {adecb}

{aebdc} {aebcd} {aedbc} {aedcb} {aecbd} {aecdb}

(and finally, 5 options for the first element, and for each of these there are 24 arrangements of the final 4 as we've shown above for all the elements beginning with a):

---

{bacde} {baced} {badce} {badec} {baecd} {baedc}

{bcade} {bcaed} {bcdae} {bcdea} {bcead} {bceda}

{bdace} {bdaec} {bdcae} {bdcea} {bdeac} {bdeca}

{beadc} {beacd} {bedac} {bedca} {becad} {becda}

---

{cabde} {cabed} {cadbe} {cadeb} {caebd} {caedb}

{cbade} {cbaed} {cbdae} {cbdea} {cbead} {cbeda}

{cdabe} {cdaeb} {cdbae} {cdbea} {cdeab} {cdeba}

{ceadb} {ceabd} {cedab} {cedba} {cebad} {cebda}

---

{dabce} {dabec} {dacbe} {daceb} {daebc} {daecb}

{dbace} {dbaec} {dbcae} {dbcea} {dbeac} {dbeca}

{dcabe} {dcaeb} {dcbae} {dcbea} {dceab} {dceba}

{deacb} {deabc} {decab} {decba} {debac} {debca}

---

{eabcd} {eabdc} {eacbd} {eacdb} {eadbc} {eadcb}

{ebacd} {ebadc} {ebcad} {ebcda} {ebdac} {ebdca}

{ecabd} {ecadb} {ecbad} {ecbda} {ecdab} {ecdba}

{edacb} {edabc} {edcab} {edcba} {edbac} {edbca}



Apr 6th, 2015

So that 5! = 5 x 4 x 3 x 2 x 1 at the beginning comes from:

5 options to start (a,b,c,d,e). For each of these:

4 options for the 2nd element (all but the one you used). For each of THESE:

3 options for the 3rd element (all but the 2 you've used). For each of THESE:

2 options for element 4 (all but the 3 used in the first 3 slots). And finally for each:

1 remaining option for the final element (the last one left over).


Hence 5 x 4 x 3 x 2 x 1. Make sense?


If you have to list out long groups like this it can help to do a portion then use find/replace to change the elements. E.g. a -> 1; b -> a; 1 ->b will swap a and b in all combinations. Or you can build them in Excel by using the CONCATENATE function on 5 columns containing elements 1-5.

Apr 6th, 2015

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