GMAT Boost Challenge Question #4

[gmatboost]

Here is a GMAT Boost Challenge Question.

On GMAT Boost, I frequently leave out the answer choices on PS questions so that the focus is on the math rather than the answers. If you have any questions, let me know.

The solution to the previous challenge question is available here:
https://www.beatthegmat.com/gmat-boost-c ... 89987.html
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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?

[cans]

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....

[pemdas]

no restriction per 1 officer to give two presentations >>>
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

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?

[saketk]

Here is a GMAT Boost Challenge Question.

On GMAT Boost, I frequently leave out the answer choices on PS questions so that the focus is on the math rather than the answers. If you have any questions, let me know.

The solution to the previous challenge question is available here:
https://www.beatthegmat.com/gmat-boost-c ... 89987.html
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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?

5 officers
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

[mehrasa]

another opinion?

[studentps2011]

3 officers can be selected & designated for 3 presentations in 5P3 ways.

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

[saketk]

Greg - Can you please post the answer now

[shankar.ashwin]

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