anuttarasammyak said:
After the idea the problem is reduced to 12 items, i.e.,
4 portraits,
5 landscapes,
1 abstract paintings group,
and 2 modern art pieces. Permutation in abstract painting group 3! should be maltiplied to the answer or the reduced problem.
For the reduced problem we focus on positons of 2 modern art pieces.
Say we walk and see x pieces before a modern art, y pieces between the modern arts and z pieces after the next modernart to the end.
The number of positons allowed for portraits is
$$N(x,y,z)=(x-1)\theta(x-1)+(y-2)\theta(y-2)+(z-1)\theta(z-1)$$
where ##\theta(x)=1 ## for x>0, otherwise 0.
Thus the number of arrangements is
$$2!4!\sum_{x,y,z \in\{0,1,2,3,4,5,6,7,8,9,10\},x+y+z=10\ } \ _{N(x,y,z)}C_{4}$$
Further 6! should be multiplied for remaining 6 positions.
$$N(x,y,z)=\max(x-1,0)+\max(y-2,0)+\max(z-1,0)$$
$$=\max(x-1,0)-x+\max(y-2,0)-y+\max(z-1,0)-z+10$$
$$=10-\min(x,1)-\min(y,2)-\min(z,1)$$
Thus N={6,7,8,9}.
Let (M) denote a modern piece and (X) one of its adjacent pieces.
**Coefficient for (N=6):**
There are six possible positions for the portrait: to the left of, between, or to the right of the two blocks "XMX XMX".
The number of arrangements is the same as the number of ways of expressing 6 as the sum of three non-negative integers.
For example,
6=0+0+6=0+1+5=0+2+4= ...
The total number of possibilities is
1+2+3+4+5+6+7=28.
**Coefficient for (N=7):**
There are seven possible positions for the portrait: to the left of or between the blocks "XMX XM".
The number of arrangements is the same as the number of ways of expressing 7 as the sum of two non-negative integers.
For example,
7=0+7=1+6=...
This gives 8 possibilities.
Including the left-right reversal, there are another 8 possibilities.
There are also 8 possibilities corresponding to positions to the right and left of "XMXMX".
Thus, the total is
8+8+8=24.
**Coefficient for (N=8):**
There are eight possible positions for the portrait:
- One position between "MX XM".
- One position to the left of "XMXM".
- One position to the right of "MXMX".
- Nine possibilities corresponding to positions to the right and left of "XMMX".
Thus, the total is
1+1+1+9=12.
**Coefficient for (N=9):**
There is one possible position for the portrait, to the right of "MMX".
Including its left-right reversal, we obtain 2 possibilities.
Therefore,
28+24+12+2=66=##\frac{11(11+1)}{2}.##
This agrees with the total number of non-negative integer solutions of (x+y+z=10), confirming the count.
