Question

A bag contains tickets bearing numbers $1,2,3,...,50$. Five tickets are drown at random from the bag without replacement. Let $p_m$ denote the probability that m is the middle number of these five numbers, then

• A. ${}^{50}{C}_5{P}_3{=}^{47}{C}_2$
• B. ${}^{50}{C}_5{P}_{24}={(}^{23}{C}_2\left){(}^{26}{C}_2\right)$
• C. ${}^{50}{C}_5{P}_{48}{=}^{47}{C}_2$
• D. $p_r-p_{51-r}$ for $3\le r\le 48$

Answer 1

Answer: (A), (B)

The correct options are
A ${}^{50}{C}_5{P}_3{=}^{47}{C}_2$
B ${}^{50}{C}_5{P}_{24}={(}^{23}{C}_2\left){(}^{26}{C}_2\right)$

$P_3$ means that $3$ is the middle number. Hence, $1$ and $2$ must be there and $2$ of the remaining $47$ numbers has to be there.So, $P_3=\frac{C_2^{47}}{C_5^{50}}$.
$P_{24}$ means that $24$ is the middle number. Hence, $2$ numbers from the first $23$ numbers and $2$ of the remaining $26$ numbers has to be there.So, $P_{24}=\frac{C_2^{23}\times C_2^{26}}{C_5^{50}}$.