360 Ways (putting the letters BBNNRRQK inside squares)

[nasi112]

Say I have the letters BBNNRRQK and I want to place them inside the squares as seen in the picture above.
Each square takes one letter at each time a combination is made.

The restrictions are: One B must be in a light square while the other B is in a dark square. K must be placed between two Rs (not necessary directly).

Examples of valid combination:
RBBKQNNR
BNQNRKRB

It says that there are 360 ways to place them inside the squares. I came up with a solution, but I feel that it is wrong. Is there a better way to do that?

Let the * be an empty square. I started with this combination.

BBR****R
The remaining letters are NNKQ
There are \(\displaystyle \frac{4!}{2!} = 12\) ways to place them.

We can have the same calculation if we have:
BR****RB = 12
RB****BR = 12
R****RBB = 12
R****BBR = 12
RBB****R = 12
including
BBR****R = 12
\(\displaystyle 12 \times 6 = 72\)

Another combination is:
BBR***R*
Now K is restricted to three positions each position gives Q three different positions. So we have:
BBR***R* = 9
RBB***R* = 9
R***BBR* = 9
R***RBB* = 9
R*BB**R* = 9
R**BB*R* = 9
R***R*BB = 9
BR***R*B = 9
BR*B**R* = 9
BR***BR* = 9
RB**B*R* = 9
RB***RB* = 9
R*B**BR* = 9
R*B**R*B = 9
R**B*RB* = 9
R***BR*B = 9
\(\displaystyle 9 \times 16 = 144\)

Another combination is when the K is restricted to two positions:
BBR**R** = 6
Same as above if we change the positions of Bs, we get 16 of them. So we have:
\(\displaystyle 6 \times 16 = 96\)

Another combination is when the K is restricted to one position:
BBR*R*** = 3
Same as above:
\(\displaystyle 3 \times 16 = 48\)

Finally, combining all of them:
\(\displaystyle 72 + 144 + 96 + 48 = 360\)

 

[Dr.Peterson]

Say I have the letters BBNNRRQK and I want to place them inside the squares as seen in the picture above.
Each square takes one letter at each time a combination is made.

The restrictions are: One B must be in a light square while the other B is in a dark square. K must be placed between two Rs (not necessary directly).

Examples of valid combination:
RBBKQNNR
BNQNRKRB

It says that there are 360 ways to place them inside the squares. I came up with a solution, but I feel that it is wrong. Is there a better way to do that?

How about this:

First choose two places for the Bishops, one dark and one light. How many ways?

Next, choose any three of the remaining 6 places for RKR, and place them in that order. How many ways?

Finally, place the Queen, then put NN in the two remaining spots.

Unfortunately, my result is too large. Perhaps they are wrong. Or have you omitted a restriction?

 

[nasi112]

How about this:

First choose two places for the Bishops, one dark and one light. How many ways?

Next, choose any three of the remaining 6 places for RKR, and place them in that order. How many ways?

Finally, place the Queen, then put NN in the two remaining spots.

Unfortunately, my result is too large. Perhaps they are wrong. Or have you omitted a restriction?

Thanks a lot Dr.Peterson for this.

This 360-ways has an official name. Fischer random chess.

Enter here:

Fischer random chess - Wikipedia

en.wikipedia.org

Go down to Setup, you will see the restrictions, perhaps same as I said. Then, go to Creating Starting Positions, they gave the mathematical approach, but I did not get it.

 

[Dr.Peterson]

Thanks a lot Dr.Peterson for this.

This 360-ways has an official name. Fischer random chess.

Enter here:

Fischer random chess - Wikipedia

en.wikipedia.org

Go down to Setup, you will see the restrictions, perhaps same as I said. Then, go to Creating Starting Positions, they gave the mathematical approach, but I did not get it.

But that says that there are 960 ways, not 360; and 960 was my answer. So it was the answer that you copied incorrectly, not the rules.

Where are you getting 360???

Their derivation is similar to mine. Please try doing what I suggested, and show me your work.

 

[nasi112]

But that says that there are 960 ways, not 360; and 960 was my answer. So it was the answer that you copied incorrectly, not the rules.

Where are you getting 360???

Their derivation is similar to mine. Please try doing what I suggested, and show me your work.

It seems that I have mixed the 960 with the 360, and the whole time I was reading 960 as 360.

How about this:

First choose two places for the Bishops, one dark and one light. How many ways?

Next, choose any three of the remaining 6 places for RKR, and place them in that order. How many ways?

Finally, place the Queen, then put NN in the two remaining spots.

Unfortunately, my result is too large. Perhaps they are wrong. Or have you omitted a restriction?

Let me start with the Bishops. 15 ways?

 

[Dr.Peterson]

Let me start with the Bishops. 15 ways?

Please say what you are doing, not just a number, so we have something more to say than no. Where do you get 15?

You want to choose a dark space for one B and a light space for the other B. How do you calculate that?

 

[nasi112]

Please say what you are doing, not just a number, so we have something more to say than no. Where do you get 15?

You want to choose a dark space for one B and a light space for the other B. How do you calculate that?

I calculated like this:

Let * be any remaining pieces

BB****** 1
*BB***** 2
**BB**** 3
***BB*** 4
****BB** 5
*****BB* 6
******BB 7
B**B**** 8
B****B** 9
B******B 10
*B**B*** 11
*B****B* 12
**B**B** 13
**B****B 14
***B**B* 15
I forgot to add
****B**B 16

or just simply \(\displaystyle 4 \times 4 = 16\) as the website suggested.

Now the numbers means the number of ways RKR can be placed in the stars.

BB****** 4
*BB***** 3
**BB**** 2
***BB*** 2
****BB** 2
*****BB* 3
******BB 4
B**B**** 2
B****B** 2
B******B 4
*B**B*** 1
*B****B* 2
**B**B** 0
**B****B 2
***B**B* 1
****B**B 2

36 Ways?

 

[Dr.Peterson]

I calculated like this:

Let * be any remaining pieces

BB****** 1
*BB***** 2
**BB**** 3
***BB*** 4
****BB** 5
*****BB* 6
******BB 7
B**B**** 8
B****B** 9
B******B 10
*B**B*** 11
*B****B* 12
**B**B** 13
**B****B 14
***B**B* 15
I forgot to add
****B**B 16

or just simply \(\displaystyle 4 \times 4 = 16\) as the website suggested.

Now the numbers means the number of ways RKR can be placed in the stars.

BB****** 4
*BB***** 3
**BB**** 2
***BB*** 2
****BB** 2
*****BB* 3
******BB 4
B**B**** 2
B****B** 2
B******B 4
*B**B*** 1
*B****B* 2
**B**B** 0
**B****B 2
***B**B* 1
****B**B 2

36 Ways?

Try doing it the short way rather than using such long lists.

To place the B's, there are 4 ways to pick a dark square, and 4 ways to pick a light square, for a total of [imath]4\times4=16[/imath]. That's all you need to show.

For RKR, you just need to pick any 3 places out of the 6 that are left. Have you learned about combinations (binomial coefficients)? Use them!

 

[nasi112]

Try doing it the short way rather than using such long lists.

To place the B's, there are 4 ways to pick a dark square, and 4 ways to pick a light square, for a total of [imath]4\times4=16[/imath]. That's all you need to show.

For RKR, you just need to pick any 3 places out of the 6 that are left. Have you learned about combinations (binomial coefficients)? Use them!

Choosing any 3 out of 6 is

\(\displaystyle \frac{6!}{3!(6 -3)!} = 20\) ways

But why when doing it with the long way, I get 36 ways?

 

[Dr.Peterson]

Choosing any 3 out of 6 is

\(\displaystyle \frac{6!}{3!(6 -3)!} = 20\) ways

But why when doing it with the long way, I get 36 ways?

Clearly your way is wrong. Did you try looking for an error? It would have helped if you had written out how you got your individual numbers.

One problem is that what you listed are ways to place the Bs and RKR, not RKR alone. So your result should have been equal to 16*20 = 320, not just 20.

A bigger problem is that you appear to be counting ways to place RKR in consecutive places, which is not what is required: "K must be placed between two Rs (not necessary directly)."

A key idea is that once you have placed the Bs, you can ignore where they are located, and just count ways to place R,K,R in ******. Separating parts of the process is essential to such calculations.

 

[nasi112]

Clearly your way is wrong. Did you try looking for an error? It would have helped if you had written out how you got your individual numbers.

The error may be that I ignored the patterns when the king is not directly between the rooks (R*K**R). My individual numbers, you mean choosing 3 out of 6? It is just the combination famous formula.

One problem is that what you listed are ways to place the Bs and RKR, not RKR alone. So your result should have been equal to 16*20 = 320, not just 20.

This one I got it.

A bigger problem is that you appear to be counting ways to place RKR in consecutive places, which is not what is required: "K must be placed between two Rs (not necessary directly)."

This is what I said about my error above.

A key idea is that once you have placed the Bs, you can ignore where they are located, and just count ways to place R,K,R in ******. Separating parts of the process is essential to such calculations.

RK***R
R*K**R
R**K*R
R***KR
*RK**R
*R*K*R
*R**KR
**RK*R
**R*KR
***RKR
**RKR*
*RKR**
R**KR*
R*K*R*
RK**R*
R*KR**
RK*R**
RKR***

18 Ways

 

[AvgStudent]

I approached this problem a bit differently.

Without any restrictions, there are [imath]\dfrac{8!}{2!2!2!}= 5040[/imath] arrangments.

Next, place the BBs in opposite colors. Fix one of the B's, which left us with 7 titles, but only 4 of those satisfy the constraint.

Then, to place K between the RR's, we assign 3 titles, but only 1 of those will meet the constraint.

Finally, since all of the constraints are mutually exclusive, multiply everything together [imath]5040(4/7)(1/3) = 960[/imath].

 

[Dr.Peterson]

I approached this problem a bit differently.

Without any restrictions, there are [imath]\dfrac{8!}{2!2!2!}= 5040[/imath] arrangments.

Next, place the BBs in opposite colors. Fix one of the B's, which left us with 7 titles, but only 4 of those satisfy the constraint.

Then, to place K between the RR's, we assign 3 titles, but only 1 of those will meet the constraint.

Finally, since all of the constraints are mutually exclusive, multiply everything together [imath]5040(4/7)(1/3) = 960[/imath].

I think what you meant to say is that the constraints are independent. The terms are practically opposites, but are easily confused.

This is not an approach I would have thought of, but it does make sense (though I'd want to be very careful to be sure I haven't fooled myself). You have said, essentially, that 4/7 of all possible arrangements will have the B's on opposite colors, and that 1/3 of all arrangements will have K between the R's.

Very interesting style!

 

[AvgStudent]

I think what you meant to say is that the constraints are independent. The terms are practically opposites, but are easily confused.

This is not an approach I would have thought of, but it does make sense (though I'd want to be very careful to be sure I haven't fooled myself). You have said, essentially, that 4/7 of all possible arrangements will have the B's on opposite colors, and that 1/3 of all arrangements will have K between the R's.

Very interesting style!

Thank you @Dr.Peterson. Yes, I meant to say independent.