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JEE Advanced · Mathematics · 6. Binomial Theorem

Coefficient of x11 in the expansion of 1+x24 1+x37 1+x412 is

  1. A 1051
  2. B 1106
  3. C 1113
  4. D 1120
Verified Solution

Answer & Solution

Correct Answer

(C) 1113

Step-by-step Solution

Detailed explanation

\(2 x_1+3 x_2+4 x_3=11\)
Possibilities are \((0,1,2) ;(1,3,0) ;(2,1,1) ;(4,1,0)\)
\(\therefore\) Required coefficients
\(=\left({ }^4 C_0 \times{ }^7 C_1 \times{ }^{12} C_2\right)+({ }^4 C_1 \times{ }^7 C_3 ~\times\) \({ }^{12} C_0)+\left({ }^4 C_2 \times{ }^7 C_1 \times{ }^{12} C_1\right)+\left({ }^4 C_4\right.\left.\times{ }^7 C_1 \times 1\right)\)
\(=(1 \times 7 \times 66)+(4 \times 35 \times 1)+(6 \times 7 ~\times\) \( 12)+(1 \times 7)\)
\(=462+140+504+7=1113\)
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