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

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.

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

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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

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?

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