Thanks for your help.
Finally I found the solution to this problem for any matrix without having
to know its number of columns
I write it down in case someone else needed it.
#Creation of b matrix
ncol<-3
b <- matrix(0, 2^ncol, ncol)
for (i in 1:ncol)
b[, ncol+1-i] <- rep(rep(c(0,1),c(2^(i-1),2^(i-1))),2^(ncol-i))
b<-cbind(b,rowSums(b))
mat <- b[ ,ncol(b):1 ] ( reverse columns making last first, etc)
b[do.call("order", split(mat, col(mat))),]
--
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