#hints
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you mean how many terms there are when the thing is expanded?
Yeah
i have solved it within seconds for exam hall by putting x1+x2 as 2 terms
Only C option satified
May I know how you derived?
the stars and bars I was telling you about earlier
it's an application of the counting argument, essentially
when you expand the thing fully, you have terms of the form
$$ x_1^ax_2^bx_3^c \ldots $$
aL
in this case you got n variables and the power is 3
the sum of the powers must be equal to 3
the powers are nonnegative numbers
you've seen this before
e.g with (x+y+z+w)^6
you have terms of the form x^a y^b z^c w^d
with a+b+c+d = 6
Ohh sorry
9C3 ?
indeed
$$ \binom{n+2}{n-1} = \frac{(n+2)!}{3!(n-1)!} = \frac{n(n+1)(n+2)}{6} $$
aL