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Operations on Rational Numbers using PEMDAS Rule

For non-negat...

Question

For non-negative integers $s$ and $r,$ let
$(\begin{matrix} s \\ r \end{matrix}) = \{\frac{s!}{r! (s - r)!} \begin{array}{l} if r \leq s \\ if r > s \end{array} For positive integers m and n, let g (m, n) = \sum_{p = 0}^{m + n} \frac{f (m, n, p)}{(\begin{matrix} n + p \\ p \end{matrix})} Where for any nonnegative integer p, f (m, n, p) = \sum_{i = 0}^{p} (\begin{matrix} m \\ i \end{matrix}) (\begin{matrix} n + i \\ p \end{matrix}) (\begin{matrix} p + n \\ p - i \end{matrix}) .$
Then which of the following statements is/are TRUE?

A

$(m, n) = (n, m) f o r a l l p o s i t i v e i n t e g e r s m, n$

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B

$(m, n + 1) = (m + 1, n) f o r a l l p o s i t i v e i n t e g e r s m, n$

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C

$(2 m, 2 n) = 2 (m, n) f o r a l l p o s i t i v e i n t e g e r s m, n$

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D

$(2 m, 2 n) = ({(m, n))}^{2} f o r a l l p o s i t i v e i n t e g e r s m, n$

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Solution

The correct option is D
$(2 m, 2 n) = ({(m, n))}^{2} f o r a l l p o s i t i v e i n t e g e r s m, n$

Explanation for the correct answers:
Step1. Calculate the value of $f (m, n, p)$ :
Given,
$\begin{array}{rcl} f (m, n, p) & = & \sum_{i = 0}^{p} (\begin{matrix} m \\ i \end{matrix}) (\begin{matrix} n + i \\ p \end{matrix}) (\begin{matrix} p + n \\ p - i \end{matrix}) \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (n + i_{C_{p}}) ({}^{p + n}{{}^{C}p - i}) [∵ (\begin{matrix} n \\ r \end{matrix}) = {}^{n}{{}^{C}r}] \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (\frac{(n + i)!}{(n + i - p)! (p)!}) (\frac{(p + n)!}{(p + n - p + i)! (p - i)!}) [∵ {}^{n}{{}^{C}r} = (\frac{n!}{(n - r)! r!})] \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (\frac{(n + i)!}{(n + i - p)! (p)!}) (\frac{(n + p)!}{(n + i)! (p - i)!}) \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (\frac{1}{(n + i - p)! (p)!}) (\frac{(n + p)!}{(p - i)!}) \times \frac{n!}{n!} [Multiply and divide by n!] \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (\frac{(n + p)!}{n! (p)!}) (\frac{n!}{(n + i - p)! (p - i)!}) \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) (\frac{(n + p)!}{(n + p - p)! (p)!}) (\frac{n!}{(n - (p - i))! (p - i)!}) \\ = & \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) ({}^{{}^{n + p}C}p) ({}^{n}{{}^{C}p - i}) \\ = & ({}^{{}^{n + p}C}p) \sum_{i = 0}^{p} ({}^{m}{{}^{C}i}) ({}^{n}{{}^{C}p - i}) [Take out the constant term outside] \\ = & ({}^{{}^{n + p}C}p) [{}^{m}{{}^{C}0} {}^{n}{{}^{C}p} + {}^{m}{{}^{C}1} {}^{n}{{}^{C}p - 1} + . . . . . . . . + {}^{m}{{}^{C}p} {}^{n}{{}^{C}0}] \\ = & {}^{{}^{n + p}C}p {}^{{}^{m + n}C}p [∵ Coefficient x_{r} in {(1 + x)}^{m} {(1 + x)}^{n} = {}^{m + n}{{}^{C}r}] \\ ∴ f (m, n, p) & = & {}^{{}^{n + p}C}p {}^{{}^{m + n}C}p____(1) \end{array}$
Step2. Calculate the value of $g (m, n)$ :
$\begin{array}{rcl} g (m, n) & = & \sum_{p = 0}^{m + n} \frac{f (m, n, p)}{(\begin{matrix} n + p \\ p \end{matrix})} \\ = & \frac{\sum_{p = 0}^{m + n} ({}^{n + p}{{}^{C}p} {}^{m + n}{{}^{C}p})}{{}^{n + p}{{}^{C}p}} [S u b s t i t u t e t h e v a l u e o f f (m, n, p)] \\ = & \sum_{p = 0}^{m + n} ({}^{m + n}{{}^{C}p}) \\ = & {(1 + 1)}^{m + n} [∵ {}^{m + n}{{}^{C}o} + {}^{m + n}{{}^{C}1} + . . . . . . . + {}^{m + n}{{}^{C}m + n} = {(1 + 1)}^{m + n}] \\ ∴ g (m, n) & = & {(2)}^{m + n} \end{array}$
Step3. Check which statement is true:
$A g (m, n) = {(2)}^{m + n} = {(2)}^{n + m} = g (n, m) B . g (m, n + 1) = {(2)}^{m + n + 1} = {(2)}^{m + 1 + n} = g (m + 1, n) C . g (2 m, 2 n) = {(2)}^{2 m + 2 n} = {(2)}^{2 (m + n)} \neq g (2 (m, n)) D . g (2 m, 2 n) = {(2)}^{2 m + 2 n} = {(2)}^{2 (m + n)} = {(2^{m + n})}^{2} = {(g (m, n))}^{2}$
Hence, Option(A),(B), (D) are correct answer.

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