The pawn can't stand alone!

Let's keep the pawn aside. Now we have five pieces which can be distributed into $4$ identical boxes in $S(5,4) = 10$ ways. Here $S(n,r)$ represents the Sterling number . The pawn can be dropped later into one of the four boxes.

Hence, there are $10 \times 4$ = $\boxed{40}$ ways.

Relevant wiki: Distinct Objects into Identical Bins

In case of distributing $n$ distinct objects into $r$ identical bins, the notation can be written as the Sterling number $S(n,r)$ or $\left\{n \atop r\right\}$ .

Naturally, if $6$ objects are to be placed into $4$ identical bins, the number of combinations will equal $\left\{6 \atop 4\right\}$ .

However, since we don't want the cases with lone pawns, we have to subtract the former expression with this exception first. For this exclusion, once a lone pawn is put into a box, there will be $3$ boxes left for the other $5$ pieces. Therefore, the combinations for this equals $\left\{5 \atop 3\right\}$ .

Thus, our desired solution = $\left\{6 \atop 4\right\}\ - \left\{5 \atop 3\right\}$

And by using recurrence relations, the generalized formula for any Sterling number is: $\left\{n \atop r\right\}\ = r\left\{n-1 \atop r\right\}\ + \left\{n-1 \atop r-1\right\}$ .

In other words, $\left\{n \atop r\right\}\ - \left\{n-1 \atop r-1\right\}\ = r\left\{n-1 \atop r\right\}$ .

That is, $\left\{6 \atop 4\right\}\ - \left\{6-1 \atop 4-1\right\}\ = \left\{6 \atop 4\right\}\ - \left\{5 \atop 3\right\}\ = 4\left\{5 \atop 4\right\}$ .

Putting $5$ objects into $4$ identical bins with no empty sets means that each box contains just $1$ object with only one box having $2$ objects. Therefore, if we just consider $2$ objects as one special object, there will be exactly $4$ objects for $4$ boxes, and the number we can choose $2$ things out of $n$ for pairing equals ${ n \choose 2 }$ . Hence, $\left\{5 \atop 4\right\} = { 5 \choose 2 } = 10$ . (See generalization of putting $n$ objects into $n-1$ identical bins in the Distinct Objects into Identical Bins for more details).

Finally, $\left\{6 \atop 4\right\}\ - \left\{5 \atop 3\right\}\ = 4\left\{5 \atop 4\right\}\ = 4\times 10 = \boxed{40}$ .

As a result, there are 40 different ways to arrange the $6$ chess pieces so that the pawn won't stand alone.

First, arrange all in 6 in 4 ways : S(6,4) These contains when pawn aside too. Take the pawn to one box. Now? You're remaining with 5 pieces and 3 boxes. Arrange them : S(5,3) Apply Simple maths : S(6,4) - S(5,3) = 40