In how many ways can 10 different paintings be distributed between two

Question

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

GMATinsight · Accepted Answer

Every painting can be given away in 2 ways (either to Dave or to Mona)

Total Ways of giving away paintings  = 2*2*2*2*2*2*2*2*2*2 = 2^{10} = 1024

Unfavorable ways = Dave gets one painting (10C1) ways + Dave gets no painting (10C0) ways = 10+1 = 11

Total Favorable ways = 1024-11 = 1013

Answer: Option B

Archit3110 · Answer

Total ways the paintings can be distributed between 2 people ; 2^10 ; 1024
Since we need to determine Dave gets atleast 2 paintings so 
assume Dave gets 0,1 painting i.e. 1+ 10c1 ; 11 ways

Atleast 2 paintings ; 1024-11 ; 1013
option B

Bunuel · Answer

Veritas Prep Official Solution

Dave should get at least two paintings so it means he can get 2 or 3 or 4 or more up to 10 paintings. Calculating all those cases would be tedious so this is a perfect opportunity to use ‘Total – Opposite’ method.

Total ways in which you can distribute 10 paintings between two people without any constraints: Each painting can be given away in two ways – either to Dave or to Mona. So the paintings can be distributed in 2*2*2*…*2 = 2^10 = 1024 ways

Number of ways in which Dave gets 0 paintings or 1 painting: 1 + 10C1 = 11 ways

So number of ways in which Dave gets 2 or 3 or 4 … upto 10 (i.e. at least 2 paintings) = 1024 – 11 = 1013 ways

ANSWER: B

imslogic · Answer

Dear all: How would one solve two related derivative questions:

1. In how many ways can I distribute 10 identical sweets at a party among 2 or 3 friends?

2. In how many ways can 10 identical paintings be distributed between 2 collectors - Dave and Mona - if Dave should get at least 2 paintings?

Hi  KarishmaB , your expert view is always welcomed. Thank you!

Posted from my mobile device

Shubhradeep · Answer

In case of identical items the scenario is different.
 
Suppose X denote the items and we divide them with | signs. So if you need to distribute between 2 people you put 1  |  
If you want to distribute among 3, then keep 2 |'s like   XX | XX | X    
Similarly, for distribution among n people, keep  n-1 |'s

Now, there's a beautiful explanation to figure out the number of ways in which n identical items can be distributed among r people :-

Suppose n=6 and r=3
So, we shall put n X's ( 6 X's in this case) in different combinations between the (r-1) |'s ( 2 |'s)
So, there are some ways which we can think of right? Like:-
 XX | XX | XX
X | XXX | XX
XXX |  | XXX 
Empty space means that person got nothing

But, in each possiblity, notice that the total number of X's and |'s is always constant - which is 6 X's and 2 |'s - so, total 8 items - within which there are 6 identical items of one type (X) and 2 identical items of second type (|'s)

Now, we know how to arrange those 8 items in such case, and number of ways of arranging them would be =  \frac{8!}{6!*2!} 
Which is in turn =  8 C2  = (n+r-1)C(r-1)

So, if your question is - number of ways 10 identical items can be distributed among 2 people -- it is  (10+2-1) C(2-1) = 11C1 = 11

If Dave has to get at least 2 items that makes the equation like :
(d+2)+m = 10  
=> d+m = 8
=> so no. of ways =  (8+2-1) C(2-1) 
 = 9C1 = 9 ways.

Now you can solve your 1st queation:
In how many ways can I distribute 10 identical sweets at a party among 2 or 3 friends?
 For 2 friends = 11C1 ways = 11 ways
For 3 friends =  (10+3-1)C(3-1) = 12C2 = 66 ways

Saya12345 · Answer

Why not the following way:
Assign 2 paintings to D - this can be done in 10C2 ways
Remaining 8 can be distributed in 2^8 ways.
Total = 10C2*2^8= 45*256=11520

Bunuel · Answer

That number would have duplicates. Consider a case where Dave gets, say, three paintings. If you select two paintings for Dave, say #1 and #2, using 10C2, and then select another painting, say #3, with 2^8, Dave would end up with paintings #1, #2, and #3. However, you'd get the same set of paintings for Dave if you select #1 and #3 with 10C2 and then #2 with 2^8. This results in double counting.

Hope it's clear.

bumpbot · Answer

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In how many ways can 10 different paintings be distributed between two

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