In how many ways can 10 different paintings be distributed between two collectors – Dave and Mona – if Dave should get

[BTGmoderatorDC]

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

[regor60]

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