Perm & Comb logical approach

[k4kar]

i have provided the various approaches generally taken by students in solving P&C problems. Plz elaborate on the exact "problem type - logic" approach to be followed.( since most of the reasoning seems to be logical!) ** WHY DO WE THINK THE WAY WE DO(esp the wrong way :p) **

5 security officers arrive at a school for a presentation. They will give a total of 3 demonstrations, in a predetermined order. Each of the demonstrations will be given by exactly 1 of the security officers. If no security officer gives all three demonstrations, in how many different orders could the officers give demonstrations?

ANSWERS 1)
the order of demonstrations is fixed. d1,d2,d3.
1st can be given by any one of the 5 officers = 5 ways
2nd -> 5 ways
3rd -> 5 ways
Total = 5*5*5 = 125 (this includes the case when one officer gives all 3 demonstrations)
Thus 125-5 = 120....

ANSWER 2)
1 presentation by 1 officer, p(5,3)=5*4*3=60
2 presentations by 1 officer, 5*5*4=100
in total 160 orders

ANSWER 3)
we need to choose 3 officers out of 5 who will give the demonstration.
This can be done in 5C3 ways = 10 ways
let the officers chosen be A,B and C.
Demonstration be D1,D2,D3
A can choose one out of (D1,D2 & D3) in 3 ways
B can choose in 2 ways
C can choose in 1 way
total 6
Answer = 10*6 =60 ways

ANSWER 4)
Hence, the solution could be,
5P3 = 5!/(5-3)! = 60 ways

ANSWER 5)
Picking up 3 profs from 5 - 5C3 = 10
3 profs and 3 demos - 3! ways = 6
Total = 6*10 = 60
Picking up 2 profs from 5 - 5C2 = 10
arrangements of demos - again 3! = 6
Total = 10*6 =60
Together, 60+60 = 120 IMO

[jnelson0612]

What is the original source of the question?

[k4kar]

from a website " gmat booster".

[RonPurewal]

from a website " gmat booster".

hi,
please provide a working link -- ideally to this specific problem, but, if not, to this "gmat booster" website in general. (i tried "gmatbooster.com" and similar things, but no actual websites came up.)
sorry for the inconvenience, but we need to verify the source of a problem (in order to avoid copyright transgressions).

thanks.

[RonPurewal]

also -- when you post a link, please try to rephrase the question you're asking. i went back and read the original post, and, honestly, i can't tell what you are really asking.

[k4kar]

My apologies Jamie and Ron!
Got the spelling wrong...(its Q4 in the below link)
its " http://blog.gmatboost.com/category/challenge-questions/"

My question:

1)what is the correct answer and the most efficient method
to solve
2)And what is wrong with the approach (which seem logical too!) in other methods?


=======================================
P.S:the sample answers were provided by students in a forum

[jlucero]

The first solution is by far the most efficient. Understand that without the restriction that no officer can deliver all three presentations, there are three possibilities:

A) Three different officers (60 ways)
B) Two different officers (60 ways)
C) One officer (5 ways)

The most elegant solution is to recognize that all three options will result in 5 * 5 * 5 = 125 total possible combinations of officers that can be selected. So "Total - C" = combinations where one officer is not doing all three presenations.

You can also find "A + B" to solve this equation, but this is where each of the wrong methods falter. Some try to combine steps A + B together (can't do that, these are separate situations), others fail in the math in solving for B (don't take in the different ways that one officer can speak twice)

Calculating A (3 different officers):
X, Y, Z
5 * 4 * 3 = 60 combinations (don't divide here because being selected first is different than being selected second)

Calculating B (2 different officers):
Here's the tricky part- these 2 different officers can show up in 3 different arrangements:
X, X, Y
X, Y, X
Y, X, X

For each of these options, there are 5 different people to choose to be the first officer, and 4 different people to choose for the second officer. So the calculation for each arrangement would be:
X, X, Y = 5 * 1 * 4 = 20
X, Y, X = 5 * 4 * 1 = 20
Y, X, X = 5 * 4 * 1 = 20

Adding possibilities A + B = 60 + 60 = 120

[k4kar]

Great! Thank you Joe! A fresh pair of eyes is helpful indeed!
thanks again!

[tim]

:)