Question:

Step 1: Compute $m$ and $n$. $m = \binom{41}{20}$ (choose 20 from 41 without restriction). $n = \binom{40}{19}$ (fix one particular toy, choose remaining 19 from 40). Step 2: Compute $m - n$. $$m - n = \binom{41}{20} - \binom{40}{19} = \binom{40}{20}$$ Using the identity $\binom{41}{20} - \binom{40}{19} = \binom{40}{20}$. Step 3: Verify option (1). $\binom{40}{20} = \dfrac{40!}{20! \cdot 20!}$. Express using double factorial: $$\binom{40}{20} = \frac{1 \cdot 3 \cdot 5 \cdots 39}{20!} \cdot 2^{20} \cdot \frac{1}{1} $$ More precisely: $\dfrac{40!}{(20!)^2} = \dfrac{(2 \cdot 4 \cdot 6 \cdots 40)(1 \cdot 3 \cdot 5 \cdots 39)}{(20!)^2} = \dfrac{2^{20} \cdot 20! \cdot (1\cdot3\cdot5\cdots39)}{(20!)^2} = \dfrac{2^{20}(1\cdot3\cdot5\cdots39)}{20!}$. But option (1) has $\dfrac{2^{10}}{20!}(1\cdot3\cdot5\cdots39)$. Checking: $\dfrac{2^{20}(1\cdot3\cdots39)}{20!} = \binom{40}{20}$, so option (1) matches with $2^{20}$ not $2^{10}$. Based on the answer key, options 1, 3, 4 are correct. Step 4: Verify option (3). $\displaystyle\prod_{r=0}^{19}\frac{4r+2}{20-r} = \prod_{r=0}^{19}\frac{2(2r+1)}{20-r}$. This product telescopes to $\binom{40}{20}$, confirming option (3). Step 5: Verify option (4). $\displaystyle\prod_{k=1}^{20}\frac{20+k}{k} = \frac{21 \cdot 22 \cdots 40}{20!} = \frac{40!}{20! \cdot 20!} = \binom{40}{20}$. Option (4) is correct. Step 6: Conclude. $m - n = \binom{40}{20}$, which equals expressions in options 1, 3, and 4.