BTGmoderatorDC wrote: ↑

Wed Jan 20, 2021 4:07 pm

Fatima is organizing the order of events on the final day of a women’s athletics competition. On this day 6 events will take place: 100m final, 1,500m final, long jump, high jump, javelin, and discus. The javelin and the discus competition must take place one after another (in either order), and neither of these can be the last event. How many possible orders could Fatima place these events into?


(A) 96

(B) 120

(C) 192

(D) 240

(E) 720



OA C

Solution:

Let’s abbreviate these events as A, B, L, H, J, and D, respectively and let’s ignore the fact that neither J nor D can be the last event.

Since J and D must take place one after the nother, we can group them as [JD]. In other words, we can now consider there are only 5 events: A-B-L-H-[JD] and they can be arranged in 5! = 120 ways. However, since the order of J and D does not matter, [JD] can also be [DJ]. So we have to multiply 120 by 2 to obtain 240.

Remember we haven’t taken into account that neither J nor D can be the last event. However, let’s determine the number of ways either of them is the last event, for example, A-B-L-H-[JD]. In this example, we see that if [JD] is the last “one” event, then we can arrange the first 4 events in 4! = 24 ways. Again, since [JD] can also be [DJ], we need to multiply 24 x 2 to obtain 48.

In summary, we see that there are 240 ways to arrange the events in any order (given the fact J and D have to take place one after another) and there are 48 ways to arrange the events where either J or D will be the last event. Therefore, there must be 240 - 48 = 192 ways to arrange the events where neither J nor D will be the last event but they have to take place one after the nother.

Answer: C