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 ${\left(1-x\right)}^{-3}\left(1+x^{22}-2x^{11}\right)\left(1-x^{10}\right)$
$=$ coeff. of $x^{10}$ in ${\left(1-x\right)}^{-3}\left(1+x^{22}-2x^{11}-x^{10}-x^{32}+2x^{21}\right)$
$=$ coeff. of $x^{10}$ in ${\left(1-x\right)}^{-3}\left(1-x^{10}-2x^{11}+2x^{21}+x^{22}-x^{32}\right)$
$=$ coeff. of $x^{10}$ in ${\left(1-x\right)}^{-3}-$ coeff. of $x^0$ in ${\left(1-x\right)}^{-3}$
${=}^{10+3-1}{C}_{10}-1{=}^{12}{C}_{10}-1=\frac{12\times 11}{2}-1=65$