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The coefficient of $x^{3}$ in the expansion of $(1 + x)^{3} (2 + x^{2})^{10}$ is.

1. $2^{14}$
2. $31$
3. $\left ( \frac{3}{3} \right ) + \left ( \frac{10}{1} \right )$
4. $\left ( \frac{3}{3} \right ) + 2\left ( \frac{10}{1} \right )$
5. $\left ( \frac{3}{3} \right ) \left ( \frac{10}{1} \right ) 2^{9}$

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$(1+x)^3 = (1+x^{3}+3x+3x^{2})$

and $(2+x^{2})^{10} = _{0}^{10}\textrm{C}*2^{0}*(x^{2})^{10} + _{1}^{10}\textrm{C}*2^{1}*(x^{2})^{9} + ................ + _{9}^{10}\textrm{C}*2^{9}*(x^{2})^{1} + _{10}^{10}\textrm{C}*2^{10}*(x^{2})^{0}$

So , coefficient of $x^{3} = _{10}^{10}\textrm{C} * 2^{10} + 3 * _{9}^{10}\textrm{C}*2^{9} = 2^{9} (32) = 2^{14}$

As here we need to multiply last term of second expansion with first term of first coefficient ( x3 ) and 3x with x2 in the second expansion.

selected

it should be 2^14

equivallent exp:(1+x)3210(1+x2/2)10

we can get x^3 in the expansion if (1+x),(1+x) and (1+x)  is multiplied together or one of the (1+x)  block multiplied with one of the (2+x^2)  block

so,coff of x^3= (3c3 + 3c1*10c1*1/2).*2^10  =  (1+3*10*1/2)*2^10  =2^14

(1+x)3 (2+ x2 )10

Lets Expand this ->

(1+x)(1+x) (1+x) (2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )(2+ x2 )

If you see here there are two ways of getting x3

Way 0 :->Choose x from each of (1+x)3 In that case we need to choose 2 from each of (2+ x2 ). Here we get 210 part of sum.

Way 1-> Here we can choose (1+x) from (1+x)3 . Which we can do in 3 ways (3 choose 1) . Then we need to choose 1 xfrom (2+ x2 )10 , Which we can do in 10 way ( 10 choose 1)

Total ways we can choose x3 using way 1 is 30 =(10* 3)

Total sum of coeffient in way 1 = 29 * 30

After calculation we get = 210 + 29 * 30 => 16384 => 214 => Answer A)

edited

Nice solution by using Combinatorics concept.

Just small correction in this line "Total ways we can choose x3 using way 1 is 30 ! (10* 3)" .

It should be 30 not 30! .