GMAT Boost Challenge Question #4

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gmatboost: Master | Next Rank: 500 Posts; Posts: 312; Joined: Tue Aug 02, 2011 3:16 pmFollowed by: 33 members; Location: New York City; Thanked: 130 times; GMAT Score:780

by gmatboost » Wed Sep 14, 2011 9:41 am

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?

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Source: Beat The GMAT — Problem Solving | Wed Sep 14, 2011 9:41 am

cans: Legendary Member; Posts: 1309; Joined: Mon Apr 04, 2011 5:34 amFollowed by: 123 members; Location: India; Thanked: 310 times; GMAT Score:750

by cans » Wed Sep 14, 2011 9:50 am

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....

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pemdas: Legendary Member; Posts: 1084; Joined: Fri Apr 15, 2011 2:33 pmFollowed by: 21 members; Thanked: 158 times

by pemdas » Wed Sep 14, 2011 10:09 am

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

gmatboost wrote: 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?

Success doesn't come overnight!

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saketk: Legendary Member; Posts: 608; Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members; Thanked: 37 times

by saketk » Wed Sep 14, 2011 12:34 pm

gmatboost wrote: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
------

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

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mehrasa: Master | Next Rank: 500 Posts; Posts: 279; Joined: Fri Nov 05, 2010 5:43 pmFollowed by: 1 member; Thanked: 15 times

by mehrasa » Mon Sep 19, 2011 10:49 am

another opinion?

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studentps2011: Senior | Next Rank: 100 Posts; Posts: 40; Joined: Fri Jul 15, 2011 5:37 am; Thanked: 3 times

by studentps2011 » Mon Sep 19, 2011 11:14 am

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

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

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saketk: Legendary Member; Posts: 608; Joined: Sun Jun 19, 2011 11:16 amFollowed by: 8 members; Thanked: 37 times

by saketk » Fri Oct 14, 2011 4:03 am

Greg - Can you please post the answer now

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shankar.ashwin: Legendary Member; Posts: 966; Joined: Sat Jan 02, 2010 8:06 amFollowed by: 21 members; Thanked: 230 times

by shankar.ashwin » Fri Oct 14, 2011 12:54 pm

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

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