Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
June 2021 p51 q3
2685
(a) How many different arrangements are there of the 8 letters in the word RELEASED?
(b) How many different arrangements are there of the 8 letters in the word RELEASED in which the letters LED appear together in that order?
(c) An arrangement of the 8 letters in the word RELEASED is chosen at random. Find the probability that the letters A and D are not together.
Solution
(a) The word RELEASED consists of 8 letters where E is repeated twice. The number of different arrangements is given by \(\frac{8!}{2!}\). Calculating this gives \(\frac{40320}{2} = 6720\).
(b) Treat LED as a single unit. This gives us 6 units to arrange: LED, R, E, A, S, E. The number of arrangements is \(\frac{6!}{2!}\) because E is repeated. Calculating this gives \(\frac{720}{2} = 360\).
(c) First, calculate the total number of arrangements where A and D are together. Treat AD as a single unit, giving us 7 units: AD, R, E, L, E, S, E. The number of arrangements is \(\frac{7!}{3!}\) because E is repeated. Calculating this gives \(\frac{5040}{6} = 840\). The probability that A and D are not together is \(1 - \frac{840}{6720} = \frac{3}{4}\) or 0.75.
