Five Rings Online Math Assessment – Guide + Free Samples [2026]

This question can be modelled by binomial distribution:

The probability of each die roll being odd is naturally ½, so:

• p = q = (1 – p) = ½.

Additionally we know:

• n = 6
• k = 4

Sample Question 2

How many ways are there to arrange 5 distinct books on 3 shelves, so that no shelf is left empty? The order of the books on the shelf doesn’t matter.

Step 1 – Total Combinations

Since there are no restrictions on the arrangements, and no placement depends on the previous ones, the number of total possible arrangements is 3⁵ = 243.

Step 2 – Combinations with an Empty Shelf

If we choose one shelf to be empty, we have 2 shelves left to arrange the books. The number of ways to choose which shelf is empty is, naturally, 3.

Therefore, the possible number of ways of arranging the 5 books on the two remaining shelves is therefore: 3·2⁵ = 96

Step 3 – Consider Double Count

The 96 cases also include cases where two shelves are empty (i.e., all books go to the same shelf). But we have subtracted each such case twice.

For instance, the case “all books are on shelf A” (i.e., shelves B and C are empty) appears twice:

• Once in the list of “shelf B is empty”.
• Once in the list of “shelf C is empty”.

There are three such cases (one for each shelf): 3·1⁵ = 3

Step 4 – Subtract and Add

The total number of possible arrangements is 243 – 96 + 3 = 150

However, it can be safely assumed that an answer of 147 (243 – 96), while not 100% accurate, is an acceptable answer which demonstrates the ability to correctly model and simplify the problem.

Bonus: Solving with Inclusion-Exclusion

Steps 2-4 (the number of “bad” cases) can also be explained with the inclusion-exclusion formula for 3 sets:

∣A∪B∪C∣=∣A∣+∣B∣+∣C∣−∣A∩B∣−∣A∩C∣−∣B∩C∣+∣A∩B∩C∣

Where:

• A – Shelf A is empty
• B – Shelf B is empty
• C – Shelf C is empty

∣A∣ = ∣B∣ = ∣C∣ = 2⁵ = 32

∣A∩B∣ (shelves A and B are empty) = 1

Therefore:

∣A∩B∣ = ∣A∩C∣ = ∣B∩C∣ = 1

And, finally:

∣A∩B∩C∣ (shelves A, B, and C are empty) = 0

And so, the total number of “bad” cases is:

32 + 32 + 32 – 1 – 1 – 1 = 93

Directly calculating the length of an arc is a complicated process involving advanced integration. Even a simple function like x² would require quite a long, complex process incorporating the inverse hyperbolic function arcsinh.

However, we can come very close to the correct answer by simply looking at the chord connecting the two points on the x2 : (2,4) and (3,9).

Since x² is convex and monotone, the chord is always above the arc and hence provides a lower bound for its length.

That length is:

From the plot above, we can see that the actual length of the arc is very close to that, but let’s consider an upper bound as well.

Since x² is monotone, the lower bound is ∆x + ∆y = 1 + 5 = 6.

So, if your answer is anywhere between 5.1 and 6, you got it right, but the closer you are to 5.1, the better.

Note: If the actual length of the arc interests you, you can check it out here.

We’ll use two methods to solve this question.

Method #1 – Sum of an Arithmetic Progression

All dual integers 1-100 are an arithmetic progression with a₁ = 2, a₅₀ = 100, n = 50.

So:

Similarly:

So, S_O – S_D = 2500 – 2550 = –50

Method #2 – “Pairs”

Let’s look at the first two pairs of numbers.

Odd: 1 + 3 = 4

Dual: 2 + 4 = 6

Odd – Dual = –2

This pattern goes on for each and every pair:

[(5 + 7) – (6 + 8)] = [(9 + 11) – (10 + 12)] = …. = –2

Since there are 25 pairs, the total difference is 25 x (–2) = –50