Find the value of \({}^{75}P_2-{}^{75}C_2\).
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 2775
\({}^{n}P_r\) counts arrangements (order matters) and \({}^{n}C_r\) counts selections (order does not matter); each selection of \(r\) things can be arranged in \(r!\) ways, so \({}^{n}C_r={}^{n}P_r\div r!\).
Arrangements of 2 out of 75: \({}^{75}P_2=75\times74=5550\)
Selections of 2 out of 75: \(\displaystyle {}^{75}C_2=\frac{5550}{2!}=2775\)
\({}^{75}P_2-{}^{75}C_2=5550-2775=2775\)
Shortcut:
Since \(2!=2\), \({}^{n}P_2=2\times{}^{n}C_2\), so \({}^{n}P_2-{}^{n}C_2={}^{n}C_2\)
So the answer is just \(\displaystyle {}^{75}C_2=\frac{75\times74}{2}=2775\)