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
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}
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 15.
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