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Joey is planning to throw a small party...

by BTGmoderatorLU » Thu Oct 26, 2017 6:01 am
Joey is planning to throw a small party and for that he needs to do the following tasks:

I. Decide guest list.
II. Send invites.
III. Arrange catering.
IV. Decide the venue.
V. Rent a music system.

The tasks can be done in any order as long as the invites are sent after the guest list is decided. In how many ways can Joey do the five tasks listed above?

A. 24
B. 48
C. 60
D. 120
E. 240

The OA is C.

I don't have it clear. Can any expert explain this PS question for me please? Thanks.
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by [email protected] » Thu Oct 26, 2017 10:41 am
Hi LUANDATO,

We're told that there are 5 tasks to be completed. IF there were no 'restrictions', then there would be 5! = (5)(4)(3)(2)(1) = 120 different possible 'orders' for those 5 tasks. However, there IS a 'restriction' - the invites can only be sent AFTER the guest list is determined. In the original 120 possible orders, the 5 tasks are done one-at-a-time, so HALF of the options had the invites sent BEFORE the guest list and HALF of the options had the invites sent AFTER the guest list. Thus, there are (1/2)(120) = 60 possible orders in which the invites are sent after the guest list.

Final Answer: C

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by GMATGuruNY » Fri Oct 27, 2017 4:21 am
LUANDATO wrote:Joey is planning to throw a small party and for that he needs to do the following tasks:

I. Decide guest list.
II. Send invites.
III. Arrange catering.
IV. Decide the venue.
V. Rent a music system.

The tasks can be done in any order as long as the invites are sent after the guest list is decided. In how many ways can Joey do the five tasks listed above?

A. 24
B. 48
C. 60
D. 120
E. 240
Alternate approach:

Let the 5 tasks = A, B, C, D and E.
To complete the tasks, Joey must create a schedule with 5 positions.
Since task A (deciding a guest list) must be completed before task B (sending invites), A must appear TO THE LEFT OF B in the schedule.

Number of options for C = 5. (Any of the 5 positions in the schedule.)
Number of options for D = 4. (Any of the 4 remaining positions in the schedule.)
Number of options for E = 3. (Any of the 3 remaining positions in the schedule.)
Number of options for A = 1. (Of the 2 remaining positions, the one more to the left must be occupied by A, since A must appear to the left of B in the schedule.)
Number of options for B = 1. (Only 1 position remains.)
To combine the options above, we multiply:
5*4*3*1*1 = 60.

The correct answer is C.
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