(1+x)^{3} (2+ x^{2 })^{10}
Lets Expand this ->
(1+x)(1+x) (1+x) (2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })(2+ x^{2 })
If you see here there are two ways of getting x^{3}
Way 0 :->Choose x from each of (1+x)^{3 }In that case we need to choose 2 from each of (2+ x^{2 }). Here we get 2^{10 }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 x^{2 }from (2+ x^{2 })^{10 }, Which we can do in 10 way ( 10 choose 1)
Total ways we can choose x^{3} using way 1 is 30 ! (10* 3)
Total sum of coeffient in way 1 = 2^{9 }* 30
After calculation we get = 2^{10} + 2^{9} * 30 => 16384 => 2^{14 }=> Answer A)