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Just for interest:
I have

reset()
import scipy.stats
n = 2000
p = 0.2
binom_dist = scipy.stats.binom(n,p)
bar_chart([binom_dist.pmf(x) for x in range(n)], frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))])

How could outsource the part like
Layout = frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))]?

And then

bar_chart([binom_dist.pmf(x) for x in range(n)], Layout)

Just for interest:
I have

reset()
import scipy.stats
n = 2000
p = 0.2
binom_dist = scipy.stats.binom(n,p)
bar_chart([binom_dist.pmf(x) for x in range(n)], frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))])

How could outsource the part like
Layout = frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))]?

And then

bar_chart([binom_dist.pmf(x) for x in range(n)], Layout)

Just for interest:
I have

reset()
import scipy.stats
n = 2000
p = 0.2
binom_dist = scipy.stats.binom(n,p)
bar_chart([binom_dist.pmf(x) for x in range(n)], frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))])

How could outsource the part like
Layout = frame=True, axes_labels=['$k$','$B(k,{},{})$'.format(n,float(p))]?

And then

bar_chart([binom_dist.pmf(x) for x in range(n)], Layout)