Method 1 (one variant)
32 = 1 x 32
So have a bag with 32 numbered tiles or balls in there, shake it around and pull one out.
Yawn.
Method 2 (three variants)
32 = 2 x 16 = 2 x 4 x 4 (or 4 x 2 x 4 or 4 x 4 x 2)
So have a coin (two sides) and two tetrahedra (each with four faces)
Divide your list into two batches of 16 and toss the coin to decide which one to keep. Then divide the 16 into 4 batches of 4 and use the first tetrahedron to decide which batch of 4 to keep. Then use the other tetrahedron to decide one from that final four. Remember it is the face down that "counts" on a tetrahedron. QED. You can also use the sequence tetrahedron  coin  tetrahedron or tetrahedron  tetrahedron  coin if you prefer. Noone will care.
Method 3 (three variants)
32 = 2 x 16 = 2 x 2 x 8 (or 2 x 8 x 2 or 8 x 2 x 2)
So have two coins (distinguishable, say a 10p and a 20p) and an octahedron (with eight faces)
Again, divide your list into two batches of 16 to start with and toss the coin. Then divide the 16 you have kept into two batches of eight and toss the other coin to select one of them. Finally, use the octahedron to pick one of that final eight. Simples. Alternatives are to use coin  octahedron  coin or octahedron  coin  coin in that order. Whatever you like.
Method 4 (two variants)
32 = 4 x 8 (or 4 x 8)
So have a tetrahedron and an octahedron. Or vice versa. You get the idea. Or have lost the will to live.
Method 5 (four variants)
32 = 2 x 2 x 2 x 4 (or 2 x 2 x 4 x 2 or 2 x 4 x 2 x 2 or 4 x 2 x 2 x 2)
Three coins and a tetrahedron, and you can use the tetrahedron at the first, second, third or fourth step. Do not pass Go, do not Collect £200.
Method 6 (one variant)
32 = 2 x 2 x 2 x 2 x 2
Yes, 32 is one of the powers of 2  it is 2 to the power 5, and I am going to use five distinguishable coins. The Yapp coffers have coffed, and I have the five smallest coins of the realm. The table below assigns a unique sequence of heads (H) and tails (T) outcomes to each of the 32 games.
1ST

2ND

3RD

4TH

5TH


ASHTON UNITED

v

WITTON ALBION

H

H

H

H

H

BRIGHOUSE TOWN

v

CROOK TOWN AFC

H

H

H

H

T

BURNHAM R

v

AFC SUDBURY

H

H

H

T

H

BURSCOUGH

v

RADCLIFFE BOR

H

H

H

T

T

CARLTON TOWN

v

BRIGG TOWN

H

H

T

H

H

CORSHAM TOWN

v

BRISTOL MANOR F

H

H

T

H

T

COVENTRY SPH

v

TIVIDALE

H

H

T

T

H

EVESHAM UNITED

v

STOURBRIDGE

H

H

T

T

T

FAREHAM TOWN

v

WEYMOUTH

H

T

H

H

H

GUISBOROUGH T

v

BISHOP AUCKLAND

H

T

H

H

T

HALLEN

v

HAMWORTHY U

H

T

H

T

H

HARTLEY WINTNEY

v

CAMBERLEY T

H

T

H

T

T

HASSOCKS

v

CHIPSTEAD

H

T

T

H

H

HORNDEAN

v

AFC PORTCHESTER

H

T

T

H

T

JARROW ROOFING BCA

v

MORPETH TOWN

H

T

T

T

H

MARSKE UNITED

v

ALBION SPORTS

H

T

T

T

T

MERSTHAM

v

CORINTHIAN CAS

T

H

H

H

H

NEWCASTLE T

v

SUTTON C'FIELD T

T

H

H

H

T

OSSETT TOWN

v

BAMBER BRIDGE

T

H

H

T

H

PEACEHAVEN & T

v

LEWES

T

H

H

T

T

PENRITH AFC

v

PADIHAM

T

H

T

H

H

PRESCOT CABLES

v

BUXTON

T

H

T

H

T

RUNCORN LINNETS

v

GLOSSOP NE

T

H

T

T

H

SOUTH PARK

v

HORSHAM

T

H

T

T

T

SPENNYMOOR T

v

LANCASTER CITY

T

T

H

H

H

TEAM NORTHUMBRIA

v

SCARBOROUGH ATH

T

T

H

H

T

WARRINGTON T

v

NEW MILLS

T

T

H

T

H

WELLS CITY

v

BRISLINGTON

T

T

H

T

T

WESTFIELD

v

AYLESBURY UNITED

T

T

T

H

H

WHYTELEAFE

v

HORLEY TOWN

T

T

T

H

T

WINCHESTER CITY

v

MANGOTSFIELD U

T

T

T

T

H

YATE TOWN

v

CHIPPENHAM TOWN

T

T

T

T

T

Lights, camera, action! Where am I going this week? It's random.
See you there!
Extension: It wouldn't be strictly random, but you could use one team's performance in a penalty shootout to choose from 32. There are 32 possible outcomes because each penalty is either scored or notscored in the same way that the toss of a coin is either heads or tails. You would need a little bit of small print  you would count the first five penalties only in cases where it continued into sudden death, and you would have to count a penalty not taken (because the side has already won or lost) as notscored. You could even toss a coin to decide which team's penalties to count. However, it is not strictly random as penalties are not true 50:50 events  more are scored than missed. It would be randomish.
Extension 2: more about the shapes
There are very few 3D objects (one polyhedron, plural polyhedra) which, if thrown in the air, have an exactly equal chance of landing on any one of the faces. They are very symmetrical, and of course all the faces must be the same shape. There are only 5 of them, and they are known as the "Platonic Solids". Any selfrespecting geek will have a Platonic relationship with these beauties:
In order from L to R they are:
The TETRAHEDRON  four equilateral triangles for the faces. When used as a die, it is the "face down" that counts. One in four chance for each face when thrown.
The CUBE  six squares for the faces, and the standard d6 die that we are all used to from our earliest days playing snakes and ladders. One in six chance for each face when thrown.
The OCTAHEDRON  eight equilateral triangles for the faces, you can think of it like two squarebased pyramids stuck together. One in eight chance for each face when thrown.
The DODECAHEDRON  twelve faces, each a rectangular pentagon, and a one in twelve chance for each face when thrown.
Diversion: Throwing one 12faced dodecahedron gives very different results from throwing two sixfaced cubes. With a decahedron, all 12 scores (112) have equal chance. With two cubes, the most likely outcome is a total of 7 because there are more ways of getting 7 than anything else. Namely 1&6, 2&5, 3&4, 4&3, 5&2 and 6&1. "1&6" isn't the same outcome mathematically as "6&1"  imagine the cubes to be different colours, so it could be red6 and blue1 and vice versa. The only way of getting 12 with two cubes is 6&6 and there is only a 1 in 36 chance of that happening whereas you have a 1 in 12 chance of getting 12 with the dodecahedron. You can't score 1 with two cubes  the smallest total is "1&1" making 2. For your homework, what are the chances of getting any "double" from rolling two cubes? (Like for getting Out of Jail in Monopoly). Answers via Twitter ;)
Meanwhile, the ICOSAHEDRON is a beautiful arrangement of 20 equilateral triangles, and no doubt it will feature one day in a decision for my travels. One in 20 chance for each face when thrown.
Platonic Dice pic taken from Wikipedia and licensed as follows:
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