(a) Evaluate \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\) correct to two decimal places. (b) A committee of 2 tutors and 5 pupils is to be formed among 6 tutors and 10 pupils. In how many ways can this be done if one particular tutor must be on the committee and two particular pupils must not be on the committee?
Explanation
(a) \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\) \(\frac{^{9}P_{3}}{^{15}C_{3}} = \frac{9!}{(9 - 3)!} \times \frac{3! 12!}{15!}\) = \(504 \times \frac{1}{455} = 1.10769 \) \(\frac{^{5}C_{3}}{^{3}P_{2}} = \frac{5!}{2! 3!} \times \frac{1}{3!}\) = \(10 \times \frac{1}{6} = 1.66667\) \(\therefore \frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}} = 1.10769 + 1.66667 = 2.77436\) \(\approxeq 2.77\) (2 d.p) (b) 6 tutors and 10 pupils to select 2 tutors and 5 pupils. Select one tutor, one more tutor is needed = \(^{5}C_{1} = 5 ways\) Keep aside 2 students, we have \(^{8}C_{5} = 56 ways\) = \(5 \times 56 = \text{280 ways}\).
