Question

The number of ways in which $10$ balls can be selected from $10$ identical green balls, $10$ identical blue balls and $9$ identical red balls are

• A. $63$
• B. $64$
• C. $65$
• D. $66$

Answer 1

Answer: (C)

Required number of ways $=$ coefficient of $x^{10}$ in
$\left(x^{0} + x^{1} + x^{2} + . . . . + x^{10}\right)^{2}\left(x^{0} + x^{1} + x^{2} + . . . . + x^{9}\right)$
$=$ coeff. of $x^{10}$ in $\left(\frac{1 - x^{11}}{1 - x}\right)^{2}\left(\frac{1 - x^{10}}{1 - x}\right)$
$=$ coeff. of $x^{10}$ in $\left(1 - x\right)^{- 3}\left(1 - x^{11}\right)^{2}\left(1 - x^{10}\right)$
$=$ coeff. of $x^{10}$ in $\left(1 - x\right)^{- 3}\left(1 + x^{22} - 2 x^{11}\right)\left(1 - x^{10}\right)$
$=$ coeff. of $x^{10}$ in $\left(1 - x\right)^{- 3}\left(1 + x^{22} - 2 x^{11} - x^{10} - x^{32} + 2 x^{21}\right)$
$=$ coeff. of $x^{10}$ in $\left(1 - x\right)^{- 3}\left(1 - x^{10} - 2 x^{11} + 2 x^{21} + x^{22} - x^{32}\right)$
$=$ coeff. of $x^{10}$ in $\left(1 - x\right)^{- 3}-$ coeff. of $x^{0}$ in $\left(1 - x\right)^{- 3}$
$=^{10 + 3 - 1}C_{10}-1=^{12}C_{10}-1=\frac{12 \times 11}{2}-1=65$