Obviously yes. This can be written as,

=>(x+y0^11

=>((x+y)^10)x(x+y)

in (x+y)^10, there is a x^10 term and when it gets multiplied with y it become x^10 y

when we consider (x+y)^a the expansion becomes,

(x^a) + A(x^(a-1))(y) +B(x^(a-2))(y^2)+...+D(x^(a-n))(y^n)+..+y^a

so in each and every term in the expansion summasion of the power is equal to a,

eg (a-1)+1 = a

(a-2)+2 = a

(a-n)+n = a

hence it is clear that x^10 y therm is there in the expansion

We can expand` (x+y)^11` using the binomial theorem:

`(x+y)^11 = sum_{k=0}^11 ((11),(k))x^k y^(11-k)`

We see that a term will contain` x^10` only when k=10. So we can let k=10 to find the full term in the expansion containing `x^10` :

`((11),(10))x^10y^1`

= `11x^10y`

Therefore `11x^10y` appears in the expansion of `(x + y)^11` , having a coefficient of 11.