Fatima is organizing the order of events on the final day of a womens

Question

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

(A) 96
(B) 120
(C) 192
(D) 240
(E) 720

egunckel10 · Accepted Answer

I got to the solution with a slightly different approach.

First, since Javelin (J) and Discuss (D) must take place one after the other, I also "glued" them, leaving us with 5 items to be arranged. Here I did not include the restriction for the last event:

Number of ways of arranging 5 elements is 5!= 5*4*3*2*1 =120. Since JD and DJ are different arrangement, I multiplied the result by 2: 120*2=240.

Now to take the restriction for the last event into consideration, I calculated the number of ways in which the last event could be either J or D: 4*3*2*1*2*1=48

Lastly, I subtracted the events that conflicted with the restriction from the total n° of possible arrangements: 240-48=192

BrentGMATPrepNow · Answer

Take the task of arranging the events and break it into  stages .

We’ll begin with the  most restrictive stage .

Stage 1 : Determine the order of the javelin and the discus events. 
There are only two possible arrangements: 1) javelin before discus and 2) discus before javelin 
So, we can complete stage 1 in  2  ways

Important: At this point, we're going to "glue" the discus and javelin together (in the order determined in stage 1) to become a single entity. This will ensure that the two events occur immediately after each other.

Note, we must now arrange the 5 objects representing the 6 events (4 of which are individual events, and 1 is the discus-javelin entity)
 
  Now let's deal with the other restriction (about the last event)

Stage 2 : Select the event for the last position
Since the last event can't be the discus-javelin entity, we must choose from one of the 4 individual events
So, we can complete stage 2 in  4  ways

Stage 3 : Select an event for the first position 
There are 4 objects (events) remaining. So we can complete stage 3 in  4  ways

Stage 4 : Select an event for the second position 
There are 5 objects (events) remaining. So we can complete stage 4 in  3  ways

Continuing with the same pattern,...
So we can complete stage 5 in  2  ways 
So we can complete stage 6 in  1  way

By the Fundamental Counting Principle (FCP), we can complete all 6 stages (and arrange all of the event) in  (2)(4)(4)(3)(2)(1)  ways (= 192 ways)

Answer: C

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

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hcp · Answer

Can anyone please help tell me what's wrong with my approach below:

Given the last event cannot be either javelin or discus, the no. of ways these 2 events can be arranged is:

5C2 x 2 = 20 
(5C2 is the no. of ways to pick 2 slots out of the 5 slots that aren't the last, and the multiplication by 2 is to account for the 2 scenarios of either javelin or discus preceding the other)

The remaining events can be arranged in the 4 remaining slots in 4! = 24 ways

The total no. of arrangements is thus 20 x 24 = 480 ways.

Ankur17000 · Answer

The problem is that you are choosing 2 places out of 5, but these two places may or may not be 1 after another. So instead of 20 options you would only be left with 8 once you put in the criteria for them to be one after another.

sarthak1701 · Answer

Let's call the 6 events: w, x, y, z, j and d. j and d stand for javelin and discus respectively. Now these 2 events have to happen together so let's consider them as one unit.

We need to arrange:

wx yz

There are 5!*2 = 240 ways of doing this, we are multiplying by 2 because j and d can be arranged in 2 ways.

However, these arrangements include situations when j and d are last so we have to subtract those:

wxyz

wxyz can be arranged in 4! ways and jd can be arranged in 2 ways, 4!*2 = 48.

We don't use 5! here because jd cannot change their position (except with each other)

Subtract the possibilities and we get 240 - 48 = 192.

Dooperman · Answer

5 Spaces to be filled.
4 Options for last space. (JD cant be placed here) 
4 options for 4th space.
3 options for 3rd space.
2 options for 2nd space 
1 option for 1st space.
JD can be placed in 2 ways.
4*4*3*2*1*2 = 192

kabhere · Answer

4!*2! + 4!*2!+ 4!*2!+4!*2! (fixing jd as 1,2 or 2,3 or 3,4 or 4,5)
= 192

tryorcry · Answer

F1 F2 L H J D

how many positions can JD take ? we consider JD as a group hence total positions will be 5 only

number of ways to choose 1 position from 5 positions = 5C1= 5
and since J or D cant be at last , hence , total position =4 
Now,
fixing positions with JD and DJ
hence possible Orderly arrangements = (4x3x2x1)x(2) = 48

Number of total positions they can hold = 4

So , desired arrangement = 4x48=192

Fatima is organizing the order of events on the final day of a womens

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Fatima is organizing the order of events on the final day of a womens

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