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 both collectors should get an even number of paintings? (All paintings should be given away.)

A) 128
B) 256
C) 420
D) 512
E) 1024

chetan2u · Accepted Answer

A very short and quick method would be.

There are 10 different paintings and they are to be distributed amongst 2 collectors.
That means each painting has two ways to be distributed, so all 10 paintings will have  2^{10}  ways

As 10 is combination of either both odd or both even. That is, if one gets odd, the other also gets odd, and if one gets even number, the other too gets even number.

So half of 2^{10} will have a combination of odd numbers and half, even numbers. =>  \frac{2^{10}}{2}=2^9=512

D

dabaobao · Answer

Official Solution  
Credit: Veritas Prep

Paintings can be distributed in the following ways:

0, 10 – One person gets 0 paintings and the other gets 10

2, 8 – One person gets 2 paintings and the other gets 8

4, 6 – One person gets 4 paintings and the other gets 6

You will need to calculate each one of these ways and then add them. Note that the ‘Total – Opposite’ method does not work here because finding the number of ways in which each person gets odd number of paintings is equally daunting.

Case 1: 0, 10

One person gets 0 paintings and the other gets 10. This can be done in 2 ways – either Dave gets all the paintings or Mona gets them.

Case 2: 2, 8

One person gets 2 paintings and the other gets 8. Select 2 paintings out of 10 for Dave in 10C2 = 45 ways. Mona could also get the 2 selected paintings so total number of ways = 45*2 = 90 ways

Case 3: 4, 6

One person gets 4 paintings and the other gets 6. Select 4 paintings out of 10 for Dave in 10C4 = 210 ways. Mona could also get the 4 selected paintings so total number of ways = 210*2 = 420

Total number of ways such that each person gets even number of paintings = 2 + 90 + 420 = 512 ways

But 512 is 2^9 – in form, suspiciously close to 2^10 we used in question 2 above. Is there some logic which leads to the answer 2^(n-1)? There is!

You have 10 different paintings. Each painting can be given to one of the 2 people in 2 ways. You do that with 9 paintings in 2*2*2… = 2^9 ways. When you distribute 9 paintings, one person will have odd number of paintings and one will have even number of paintings (0 + 9 or 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5).

The tenth painting needs to be given to the person who has the odd number of paintings so you give the tenth painting in only one way. This accounts for all cases in which both get even number of paintings.

Total ways = 2^9 * 1 = 512

ANSWER: D

MBAB123 · Answer

Hey  dabaobao , did you happen to come across the answer to the question when the number of paintings were  odd . (I'm assuming you read this question in  VeritasKarishma  thread of combinatorics). It would be 512 as well right? Considering after 2^9 distribution, the last painting would be given to the person having an even of paintings (i.e, only one option). Also, generally Odds and evens tend to split up equally.

Gknight5603 · Answer

0 and 10
2 and 8
4 and 6
6 and 4
8 and 2
10 and 0

2(10C0+10c2+10c4)= 512

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Mugdho · Answer

is zero a even number ? zero is divisible by both 2 and 3..how can it be a even number?

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chetan2u · Answer

6 is also divisible by both 2 and 3, but it is even.

Divisible by 2 means even. And not divisible by 2 means it is odd. No other rule for even and odd

gmatchile · Answer

10 paintings must be divided between two people (Dave and Mona), in such a way that the amount that each person receives must always be even and 10 paintings are always distributed.

Let's see:

CASE 1
Dave : 0 paintings
Mane: 10 paintings

or exclusive (both events cannot occur simultaneously)

Dave: 10 paintings
Mane: 0 paintings

Note: Zero is even.

CASE 2
Dave: 2 paintings
Mane: 8 paints

or exclusive

Dave: 8 paintings
Mane: 2 paints

CASE 3
Dave: 4 paintings
Mane: 6 paints

or exclusive

Dave: 6 paintings
Mane: 4 paints

There are no more cases.

Let's analyze the three cases.

In each case, there are two situations, both of which must be accounted for.

CASE 1

Situation 1: Dave receives 0 frames and Mane receives 10 frames.
Situation 2: Mane receives 0 frames and Dave receives 10 frames.

Both situations must be considered when counting.

We will focus to count when everyone receives zero.

Since from a universe of 10, we choose zero, it is a combination or a permutation of 10 over zero.
What do you think?
Is it a combination or a permutation?

Very good!

It is a combination. This is clearer when we choose to deal a non-zero pair, for example 2. Here all the arrangements that can be made with 2 (2!) must be considered as a single case, because it does not matter how these are delivered. two pictures.

CASE 1:
We will attend when everyone (Dave or Mona) receives zero paints.

10C0 : Combination of 10 over zero.

Since in case 1, Dave can receive zero paintings or Mona can receive zero paintings, then

Case 1: 2x(10C0) = 2 x ((10!)/((10-0)!x0!)
 = 2 x ((10!)/((10!)x1)
 = 2 x 1
 = 2
 
Eye 0!=1

Keep in mind: In CASE 1, When Dave receives 0 frames simultaneously, Mona receives 10 frames, that is, if we count, we consider what Dave receives or what Mona receives, but not both, since both are the same situation when counting.

CAS0 2:
We will attend when everyone (Dave or Mona) receives 2 paintings.

2 x ((10!)/(10-2)!x2!)
2x ((10x9x8!)/(8)!x2!)
2x ((10x9)/(2x1))
90

CASE 3:
We will attend when everyone (Dave or Mona) receives 4 paintings.

2 x ((10!)/((10-4)!x4!))
2x ((10x9x8x7x6!)/(6!x4!))
2x ((10x9x8x7)/4!)
2x((10x9x8x7)/(4x3x2x1))
2x (10x3x7)
420

Then CASE 1 + CASE 2 + CASE 3= 512

Answer D

CHALLENGE:
What would happen if? :
- in CASE 1, we focus when both Dave and Mona receive 10 paintings.
- in CASE 2, we focus when both Dave and Mona receive 8 paintings.
- in CASE 3, we focus when both Dave and Mona receive 6 paintings.

Would the 512 answer change?

bumpbot · Answer

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

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