P&C examples : finding no of ways

Question

The number of ways can 10 balls be chosen from an urn containing 10 identical green balls , 5 identical yellow balls and 3 identical blue balls are_______ .

Comments

https://math.stackexchange.com/questions/210302/how-many-ways-balls-can-be-selected

see the answer by Brian....it has a nice approach :)

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I tried this approach but answer that is coming using this approach is 23. I exactly replicated steps followed by brian with values given in ques. kindly help me in getting correct ans. using this approach.

Answer 1

$\begin{align*} &\text{We need no of solutions of the following Eq.} \\ & x_1 +x_2 +x_3 = 10 \\ \\ &\text{Where } 0\leq x_1\leq 10 \text{ , }0\leq x_2\leq 5 \text{ , }0\leq x_3\leq 3 \\ \\ \hline \\\\ &\text{Using generating function} \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left ( 1+x+x^2+ \dots + x^{10} \right )*\left ( 1+x+x^2+ \dots + x^{5} \right )*\left ( 1+x+x^2+ x^{3} \right ) \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \frac{1-x^{11}}{1-x} \right ]*\left [ \frac{1-x^{6}}{1-x} \right ]*\left [ \frac{1-x^{4}}{1-x} \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \left ( 1-x^{11} \right )*\left ( 1-x^{6} \right )*\left ( 1-x^{4} \right ) \right ]*\left [ \left ( \frac{1}{1-x} \right )^3 \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [ \left ( 1-{\color{red}{\bf x^{11}}} \right )*\left ( 1-x^4-x^6+x^{10} \right ) \right ]*\left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ \\ &\text{drop }x^{11} \text{ term} \\ \\ \end{align*}$

$\begin{align*} \\ \\ &\text{Now ,} \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\left [1-x^4-x^6+x^{10} \right ]*\left [ \sum_{k=0}^{\infty}\binom{3+k-1}{k}x^k \right ] \\ &{\color{red}{\bf \left [ x^{10} \right ]}}:\begin{cases} & {\color{blue}{\bf +}}\;1*\binom{12}{10} = +66 \;\;\;\;\; {\color{blue}{+x^0 \text{ and k=10}}} \\ & {\color{blue}{\bf -}}\;1*\binom{8}{6} \;= -28 \;\;\;\;\; {\color{blue}{-x^4 \text{ and k=6}}} \\ & {\color{blue}{\bf -}}\;1*\binom{6}{4} \;= -15 \;\;\;\;\; {\color{blue}{-x^6 \text{ and k=4}}} \\ & {\color{blue}{\bf +}}\;1*\binom{2}{0} \;= +1 \;\;\;\;\;\; {\color{blue}{+x^{10} \text{ and k=0}}} \\ \end{cases} \\ \\ &\text{Answer } = 66-28-15+1 = 24 \\ \end{align*}$

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Similar QS :

https://gateoverflow.in/114815/mathematics

Comments

Okay. I think,It will give correct in all cases, as we are just ignoring the numerator which can affect the answer.

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My point was for $r$ greater than $10$ , it will not work.

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If $ x_i $ upper limit is upto $ r$ then only that works.