Question

For non-negative integers $s$ and$r,$ let

$$\begin{gathered} \left(\begin{array}{c}\mathrm{s}\\ \mathrm{r}\end{array}\right)=\left\{\frac{\mathrm{s}!}{\mathrm{r}!\left(\mathrm{s}-\mathrm{r}\right)!}\begin{array}{l}\mathrm{if}\mathrm{r}\le \mathrm{s}\\ \mathrm{if}\mathrm{r}>\mathrm{s}\end{array}\right. \\ \mathrm{For}\mathrm{positive}\mathrm{integers}\mathrm{m}\mathrm{and}\mathrm{n},\mathrm{let} \\ \mathrm{g}\left(\mathrm{m},\mathrm{n}\right)=\sum _{\mathrm{p}=0}^{\mathrm{m}+\mathrm{n}}\frac{\mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)}{\left(\begin{array}{c}\mathrm{n}+\mathrm{p}\\ \mathrm{p}\end{array}\right)} \\ \mathrm{Where}\mathrm{for}\mathrm{any}\mathrm{nonnegative}\mathrm{integer}\mathrm{p}, \\ \mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)=\sum _{\mathrm{i}=0}^{\mathrm{p}}\left(\begin{array}{c}\mathrm{m}\\ \mathrm{i}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{i}\\ \mathrm{p}\end{array}\right)\left(\begin{array}{c}\mathrm{p}+\mathrm{n}\\ \mathrm{p}-\mathrm{i}\end{array}\right). \end{gathered}$$

Then which of the following statements is/are TRUE?

• A. $(m,n)=(n,m)forallpositiveintegersm,n$
• B. $(m,n+1)=(m+1,n)forallpositiveintegersm,n$
• C. $(2m,2n)=2(m,n)forallpositiveintegersm,n$
• D. $(2m,2n)=({(m,n))}^{2}forallpositiveintegersm,n$

Answer 1

Answer: (A), (B), (D)

$(2m,2n)=({(m,n))}^{2}forallpositiveintegersm,n$

Explanation for the correct answers:

Step1. Calculate the value of $\mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)$:

Given,

$\begin{array}{rcl}\mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)& =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left(\begin{array}{c}\mathrm{m}\\ \mathrm{i}\end{array}\right)\left(\begin{array}{c}\mathrm{n}+\mathrm{i}\\ \mathrm{p}\end{array}\right)\left(\begin{array}{c}\mathrm{p}+\mathrm{n}\\ \mathrm{p}-\mathrm{i}\end{array}\right)\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^m{}^{C}i\right)\left(n+i_{C_p}\right)\left({}^{p+n}{}^{C}p-i\right)\mathbf{[}\mathbf{\because }\left(\begin{array}{c}n\\ r\end{array}\right)\mathbf{=}{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{r}\mathbf{]}\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^m{}^{C}i\right)\left(\frac{\left(n+i\right)!}{\left(n+i-p\right)!\left(p\right)!}\right)\left(\frac{\left(p+n\right)!}{\left(p+n-p+i\right)!\left(p-i\right)!}\right)\left[\mathbf{\because }{}^{\mathbf{n}}{}^{\mathbf{C}}\mathbf{r}\mathbf{=}\mathbf{\left(}\frac{\mathbf{n}\mathbf{!}}{\left(n-r\right)\mathbf{!}\mathbf{r}\mathbf{!}}\mathbf{\right)}\right]\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^m{}^{C}i\right)\left(\frac{\left(n+i\right)!}{\left(n+i-p\right)!\left(p\right)!}\right)\left(\frac{\left(n+p\right)!}{\left(n+i\right)!\left(p-i\right)!}\right)\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^m{}^{C}i\right)\left(\frac{1}{\left(n+i-p\right)!\left(p\right)!}\right)\left(\frac{\left(n+p\right)!}{\left(p-i\right)!}\right)\times \frac{n!}{n!}\mathbf{}\mathbf{[}\mathbf{Multiply}\mathbf{}\mathbf{and}\mathbf{}\mathbf{divide}\mathbf{}\mathbf{by}\mathbf{}\mathbf{n}\mathbf{!}\mathbf{]}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}& =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^{\mathrm{m}}{}^{\mathrm{C}}\mathrm{i}\right)\left(\frac{\left(\mathrm{n}+\mathrm{p}\right)!}{\mathrm{n}!\left(\mathrm{p}\right)!}\right)\left(\frac{\mathrm{n}!}{\left(\mathrm{n}+\mathrm{i}-\mathrm{p}\right)!\left(\mathrm{p}-\mathrm{i}\right)!}\right)\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^{\mathrm{m}}{}^{\mathrm{C}}\mathrm{i}\right)\left(\frac{\left(\mathrm{n}+\mathrm{p}\right)!}{\left(\mathrm{n}+\mathrm{p}-\mathrm{p}\right)!\left(\mathrm{p}\right)!}\right)\left(\frac{\mathrm{n}!}{\left(\mathrm{n}-\left(\mathrm{p}-\mathrm{i}\right)\right)!\left(\mathrm{p}-\mathrm{i}\right)!}\right)\\ & =& \sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^{\mathrm{m}}{}^{\mathrm{C}}\mathrm{i}\right)\left({}^{{}^{\mathrm{n}+\mathrm{p}}\mathrm{C}}\mathrm{p}\right)\left({}^{\mathrm{n}}{}^{\mathrm{C}}\mathrm{p}-\mathrm{i}\right)\\ & =& \left({}^{{}^{\mathrm{n}+\mathrm{p}}\mathrm{C}}\mathrm{p}\right)\sum _{\mathrm{i}=0}^{\mathrm{p}}\left({}^{\mathrm{m}}{}^{\mathrm{C}}\mathrm{i}\right)\left({}^{\mathrm{n}}{}^{\mathrm{C}}\mathrm{p}-\mathrm{i}\right)\left[\mathbf{Take}\mathbf{}\mathbf{out}\mathbf{}\mathbf{the}\mathbf{}\mathbf{constant}\mathbf{}\mathbf{term}\mathbf{}\mathbf{outside}\right]\\ & =& \left({}^{{}^{\mathrm{n}+\mathrm{p}}\mathrm{C}}\mathrm{p}\right)\left[{}^{\mathrm{m}}{}^{\mathrm{C}}0{}^{\mathrm{n}}{}^{\mathrm{C}}\mathrm{p}+{}^{\mathrm{m}}{}^{\mathrm{C}}1{}^{\mathrm{n}}{}^{\mathrm{C}}\mathrm{p}-1+........+{}^{\mathrm{m}}{}^{\mathrm{C}}\mathrm{p}{}^{\mathrm{n}}{}^{\mathrm{C}}0\right]\\ & =& {}^{{}^{\mathrm{n}+\mathrm{p}}\mathrm{C}}\mathrm{p}{}^{{}^{\mathrm{m}+\mathrm{n}}\mathrm{C}}\mathrm{p}\left[\mathbf{\because }\mathbf{Coefficient}\mathbf{}{\mathbf{x}}_{\mathbf{r}}\mathbf{}\mathbf{in}\mathbf{}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{m}}{\mathbf{(}\mathbf{1}\mathbf{+}\mathbf{x}\mathbf{)}}^{\mathbf{n}}\mathbf{=}{}^{\mathbf{m}\mathbf{+}\mathbf{n}}{}^{\mathbf{C}}\mathbf{r}\right]\\ \therefore \mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)& =& {}^{{}^{\mathrm{n}+\mathrm{p}}\mathrm{C}}\mathrm{p}{}^{{}^{\mathrm{m}+\mathrm{n}}\mathrm{C}}\mathrm{p}\mathrm{\_\_\_\_}\left(1\right)\end{array}$

Step2. Calculate the value of $g\left(m,n\right)$:

$\begin{array}{rcl}\mathrm{g}\left(\mathrm{m},\mathrm{n}\right)& =& \sum _{\mathrm{p}=0}^{\mathrm{m}+\mathrm{n}}\frac{\mathrm{f}\left(\mathrm{m},\mathrm{n},\mathrm{p}\right)}{\left(\begin{array}{c}\mathrm{n}+\mathrm{p}\\ \mathrm{p}\end{array}\right)}\\ & =& \frac{{\displaystyle \sum _{p=0}^{m+n}}\left({}^{n+p}{}^{C}p{}^{m+n}{}^{C}p\right)}{{}^{n+p}{}^{C}p}\mathbf{[}\mathbf{S}\mathbf{u}\mathbf{b}\mathbf{s}\mathbf{t}\mathbf{i}\mathbf{t}\mathbf{u}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{v}\mathbf{a}\mathbf{l}\mathbf{u}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{f}\left(m,n,p\right)\mathbf{]}\\ & =& \sum _{p=0}^{m+n}\left({}^{m+n}{}^{C}p\right)\\ & =& {(1+1)}^{m+n}\left[\mathbf{\because }{}^{\mathbf{m}\mathbf{+}\mathbf{n}}{}^{\mathbf{C}}\mathbf{o}\mathbf{+}{}^{\mathbf{m}\mathbf{+}\mathbf{n}}{}^{\mathbf{C}}\mathbf{1}\mathbf{}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{+}{}^{\mathbf{m}\mathbf{+}\mathbf{n}}{}^{\mathbf{C}}\mathbf{m}\mathbf{+}\mathbf{n}\mathbf{=}{\left(1+1\right)}^{\mathbf{m}\mathbf{+}\mathbf{n}}\right]\\ \therefore g\left(m,n\right)& =& {\left(2\right)}^{m+n}\end{array}$

Step3. Check which statement is true:

$$\begin{gathered} \mathbf{A}\mathrm{g}\left(\mathrm{m},\mathrm{n}\right)={\left(2\right)}^{\mathrm{m}+\mathrm{n}}={\left(2\right)}^{\mathrm{n}+\mathrm{m}}=\mathrm{g}(\mathrm{n},\mathrm{m}) \\ \mathbf{B}\mathbf{.}\mathbf{}\mathrm{g}\left(\mathrm{m},\mathrm{n}+1\right)={\left(2\right)}^{\mathrm{m}+\mathrm{n}+1}={\left(2\right)}^{\mathrm{m}+1+\mathrm{n}}=\mathrm{g}\left(\mathrm{m}+1,\mathrm{n}\right) \\ \mathbf{C}\mathbf{.}\mathbf{}\mathrm{g}\left(2\mathrm{m},2\mathrm{n}\right)={\left(2\right)}^{2\mathrm{m}+2\mathrm{n}}={\left(2\right)}^{2(\mathrm{m}+\mathrm{n})}\ne \mathrm{g}\left(2\left(\mathrm{m},\mathrm{n}\right)\right) \\ \mathbf{D}\mathbf{.}\mathbf{}\mathrm{g}\left(2\mathrm{m},2\mathrm{n}\right)={\left(2\right)}^{2\mathrm{m}+2\mathrm{n}}={\left(2\right)}^{2(\mathrm{m}+\mathrm{n})}={\left(2^{\mathrm{m}+\mathrm{n}}\right)}^2={\left(\mathrm{g}\left(\mathrm{m},\mathrm{n}\right)\right)}^2 \end{gathered}$$

Hence, Option(A),(B), (D) are correct answer.