In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get at least two paintings? (All paintings should be given away.)
A) 1012
B) 1013
C) 1014
D) 1023
E) 1024
OA B
Source: Veritas Prep
In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get
This topic has expert replies

 Moderator
 Posts: 7106
 Joined: Thu Sep 07, 2017 4:43 pm
 Followed by:23 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats

 Master  Next Rank: 500 Posts
 Posts: 416
 Joined: Thu Oct 15, 2009 11:52 am
 Thanked: 27 times
Each painting has a "choice" between each of the two so the number of ways the paintings can be distributed without restrictions is
2^10 = 1024
However, this includes the cases where Dave gets 0 and 1 painting, in violation of the restriction, so these two cases need to be removed.
The case of Dave getting 0 paintings can only occur 1 way with Mona receiving all the paintings.
Dave can get only 1 painting 10 ways, since there are 10 paintings.
So a total of 11 ways must be subtracted from 1024 = 1013
2^10 = 1024
However, this includes the cases where Dave gets 0 and 1 painting, in violation of the restriction, so these two cases need to be removed.
The case of Dave getting 0 paintings can only occur 1 way with Mona receiving all the paintings.
Dave can get only 1 painting 10 ways, since there are 10 paintings.
So a total of 11 ways must be subtracted from 1024 = 1013