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when m = 2000 show by formal calculations that

i) ## \epsilon (500,199) \epsilon (1999,10) = \epsilon (500,10) ##

ii) ## \epsilon (1999,10) \epsilon (500,1999) = 0##

hence generalise for ##\epsilon (i,j) \epsilon (k,l)## and find a single equation using the kroneeker delta symbol for the generalisation

my attempt thus far:

i) rewriting this as ##\delta _{500r} \delta _{1999s} \delta _{1999r} \delta _{10s} ## then we note that ## \delta _{1999s} \delta _{1999r} = \epsilon (1999,1999) ## which gives a 0 everywhere except the 1999th ith position and 1999th jth position but I'm not sure how I can conclude that this gives ##\epsilon (500,10)##

that's all Ihave so far unfortunately