Nauseated

 The correct and highly accurate solutions presented below for your entertainment and edification  . . . .

$$\;.\hspace{20pt} \text {a)\; } 4^3 = 64 \; \hspace{15pt}\leftarrow \hspace{15pt} \small \textcolor[rgb]{0,0,0}{\text {No restrictions:}}\\$$

$$\displaystyle \text {b)\; 3} \hspace{15pt}\leftarrow \hspace{15pt} \small \text {Enter "IntegerPartitions [3]" \; into \ Wolfram \ alpha.} \

\hspace{15pt} \text { This will show the values and the symbolic partitions.}\\
\;. \hspace{35pt}\tiny \text { (You may need to use your brain a little more than normal). }\\$$

$$\;. \hspace{15pt} \displaystyle \text {c)\; 20 } \hspace{15pt}\leftarrow \hspace{15pt} \left( {\begin{array}{*{20}c} Removed \ errant \ Binomial \\ \ from \ this \ post \\ \end{array}} \right) = \underbrace{\dfrac{(N+k-1)!}{k!}}_{Dist. \ box \ Indist. \ candy }\; \hspace{15pt}| \hspace{15pt} \Text {N=4 \; k=3} \\\\$$

$$\;. \hspace{15pt} \displaystyle \text {d)\; 7 } \hspace{15pt}\leftarrow \hspace{15pt} \binom {3}{3}\; +\; \binom {3}{2}\; +\; \binom {3}{1}\; =\; 7 \;
\text {(Indist. box; Dist. candy)}\\$$

$$\;. \hspace{15pt}\displaystyle \text {e)\; 40 } \hspace{15pt}\leftarrow \hspace{15pt} \left( \dfrac{4!}{2!}\right)* \left(2 \right) \;+\; \left( \dfrac{4!}{2!}\right) \;+\; \left(4\right)\\$$

$$\text {Reasoning:}\\
\displaystyle \text {All candy in 1 of 4 boxes (4)}\\
\displaystyle \text {2 combinations (XX Y\; \&\; XY X) \;distributed to boxes in (2)(4*3) = (2)(4!/2!) \;=\;(12*2) ways;}\\
\text {and 1 candy to each box distributed to boxes in (1)(4*3*2)/2 = 4!/2! = (12) ways.} \\
\text {Total 4+24+12 = 40 ways.}$$

All this plus or minus CDD . . .

$$\tiny \textcolor[rgb]{1,,0}{\text {(Corrected error in binomial formula for soultion c)}}\\$$

The center density is 162.2 g/cc.

Long before John figures that out, he will have graduated from stealing apples to stealing cherries.

This doesn’t give any restrictions other than both the b***s and boxes are distinguishable.

Here are solutions for certain restrictions or freedoms.

If any box(N) can contain zero to (k) b***s then the solution is

$$\displaystyle N^k \ = \ {3^6} \ = \ 729$$

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If a box MUST contain (1) but can contain up to (k-N+1) b***s then the solution is

$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \\$$

This is a scripted link to Wolfram alpha

CPhill, you didn’t miss anything. In this case, you caught something – the error in the binomial formula. (Now corrected and noted).

My CDD flare-ups are probably permanent, but the contagion is contained thanks to you.

Along with correcting the errors I clarified the maths formula by making it more consistent and uniform.

The general formula is

$$\left( {\begin{array}{*{20}c} 2N \\ N \\ \end{array}} \right) = \dfrac{{(2N)!}}{{N! (2N-N)!}}\; \hspace{15pt}\leftarrow \hspace{15pt} \tiny \textcolor[rgb]{1,,0}{\text {(Corrected error in binomial formula)}}\\$$

This works for all square grids from N=1 to any finite number.

A single square is

$$\left( {\begin{array}{*{20}c} 2(1) \\ 1 \\ \end{array}} \right) = \; 2\\$$

(one unit up and one unit right, or one unit right and one unit up).

This can be extended to a non-square grid of (R * C) where R and C are rows and columns.

For this the general formula is

$$\left( {\begin{array}{*{20}c} R+C \\ C \\ \end{array}} \right)\;=\;
\dfrac{{(R + C)!}}{{R!\; C!}} \;=\; \text {unique\;paths}
\hspace{15pt}\leftarrow \hspace{15pt} \tiny \textcolor[rgb]{1,,0}{\text {(Corrected \& clarified )}}\\$$ 

I lack the skills to demonstrate a formal maths proof. Here’s the basic logic:

 In a square grid of size N when the number of units between diagonal corners is set to equal 2N, half the route will compose the X direction and half will compose the Y direction. The position of the square (X or Y) determines the position of subsequent squares (X or Y). The value of interest is the unique routes consisting of equal X and Y directions, this is found by “choosing the N positions from the 2N that are available.

(The solutions are close to the number of ways to catch CDD).

$$\tiny \textcolor[rgb]{1,,0}{\text {(Edited: Corrected errors \& clarified )}}\\$$

The above answer was used for a similar question on another forum. That grid was 3X3 instead of 6X6. (That may have been what was intended). CDD aside, the posted solution is accurate for a 3X3 grid.

However, this question as stated does posit inconsistent restrictions.

Melody is correct. There are no 6 unit solutions.

All 6-unit paths from A to B are 3 up and 3 across. Just choose 3 of the 6 (This pertains to 6-unit paths only. Not 12 unit paths, or 9-unit paths, or the path of a Christmas goose).

$$\; Solution: \\\\
\left( {\begin{array}{*{20}c} n \\ k \\ \end{array}} \right) = \dfrac{{n!}}{{k!\left( {n - k} \right)!}}\; \hspace{15pt}| \hspace{15pt} \Text {n=6 \; k=3} \\
\left( {\begin{array}{*{20}c} 6 \\ 3 \\ \end{array}} \right) \; = \; 20 \; \tezt {paths}\\$$

Oh, that’s right. The book would have to be replaced after it was selected.

It softly and suddenly vanishes away after each selection like the baker, after the boojum variety of snark selects him.

Makes it kind of difficult to select 4 books for reading, doesn’t it?

A castle (usually known as a rook) moves laterally and a bishop moves diagonally. That means that they can never both threaten each other simultaneously. So the probability that one piece threatens the other is equal to the probability that the rook threatens the bishop, plus the probability that the bishop threatens the rook.

A rook placed anywhere on the board threatens 14 squares. The probability that the rook threatens the bishop is 14/63, or 2/9.

The bishop . . .

Threatens 7 squares when placed on one of the 28 edge squares.

Threatens 9 squares when placed on one of the 20 squares one square from the edge.

Threatens 11 squares when placed on one of the 12 squares two squares from the edge.

Threatens 13 squares when placed on one of the 4 center squares.

The average number of squares threatened by a bishop is

$$\dfrac{\left(7\times28 \right ) + \left(9\times20 \right ) + \left(11\times12 \right ) + \left (13\times4 \right )}{64}\; = \; \dfrac{35}4 \\\\

\small \text {The probability that the bishop threatens the rook} \;\dfrac{35}{4\times63} = \dfrac5{36}\\\\

\text {The probability that either threatens the other } \;
\dfrac{5}{36}+ \dfrac{2}{9} \;=\; \dfrac{13}{36}\\$$

Here you can see that getting hooked by a rook is 62.5% more likely than getting buggered by a bishop.

I’m not sure where the knight came in. Maybe Cphill is taking a break from his quest.

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