|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
18 Apr 2022, 09:19
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
69% (00:55) correct
31%
(01:12)
wrong
based on 249
sessions
History
Date
Time
Result
Not Attempted Yet
A board of directors consists of 10 members. Three of those members are in leadership positions: President, Vice President, and Treasurer. If no individual can hold more than one leadership position, how many different ways are there to assign the three leadership positions?
(A) 6
(B) 24
(C) 120
(D) 540
(E) 720
(A) 6
(B) 24
(C) 120
(D) 540
(E) 720
Archit3110
Major Poster
18 Apr 2022, 10:13
Kudos
Bookmarks
GMATNinja
total ways to select 3 members ; 10c3 ;
and ways in which 3 selected people can be chosen for positions ; 3*2*1 ; 6
10c3 * 6 ; 720
option E
PyjamaScientist
Admitted - Which School Forum Moderator
Joined: 25 Oct 2020
Last visit: 04 Apr 2026
Posts: 1,125
Given Kudos: 633
Schools: Ross '25 (M$)
GMAT 1: 740 Q49 V42 (Online)
Products:
18 Apr 2022, 20:43
Kudos
Bookmarks
Counting method:
How many choices for the first place = \(10\). For second place = \(9\). For third place = \(8\). Total no. of possibilities = \(10 * 9 * 8 = 720.\)
Combinations method:
We need to choose \(3 \) people out of \(10\) people for the post of President, VP and Treasurer. Suppose we choose \(A, B,\) and \(C\) from \(10\) directors. But either one of them can take either position. So, we don't just need to choose \(3 \) people out of \(10\), we also need to jumble them among themselves as anyone can be President/VP/Treasurer among the three. Thus, it can be done in \(10_C_3\) * \(3!\) ways = \(720\).
How many choices for the first place = \(10\). For second place = \(9\). For third place = \(8\). Total no. of possibilities = \(10 * 9 * 8 = 720.\)
Combinations method:
We need to choose \(3 \) people out of \(10\) people for the post of President, VP and Treasurer. Suppose we choose \(A, B,\) and \(C\) from \(10\) directors. But either one of them can take either position. So, we don't just need to choose \(3 \) people out of \(10\), we also need to jumble them among themselves as anyone can be President/VP/Treasurer among the three. Thus, it can be done in \(10_C_3\) * \(3!\) ways = \(720\).
