Probabilistic Methods

Order matters + no repetition → permutation. Order irrelevant + no repetition → combination. Order matters + repetition allowed → sequence. Circular arrangement → circular permutation. Repeated elements present → permutation with repeats.

Order matters, no repetition — full permutation of 7 distinct objects.

Order does not matter, no repetition — combination.

Order matters (different medals), no repetition — 3-permutation.

Each position independently chooses from 10 digits — sequence with repetition.

BANANA has 6 letters: B×1, A×3, N×2. Order matters but identical letters are indistinguishable — permutation with repeated elements.

Rotations of the same arrangement are identical — circular permutation. Fix one person to remove rotational equivalence.

Comparing the five finite results (the $10^6$ PIN sequence is excluded — it dwarfs the rest).

All 8 distinct objects, order matters — full permutation.

Glue the pair into one unit, giving 7 objects to arrange. The pair can swap internally in 2! ways.

Subtract the adjacent cases from the unrestricted total.

One position is determined; freely arrange the remaining 9 questions.

The first three results share the same order of magnitude and are easy to compare directly (9! = 362880 is charted separately as it is nearly 9$\times$ larger).

When a word of length $n$ contains repeated letters with frequencies $n_1, n_2, \ldots$, the number of distinct arrangements is $\frac{n!}{n_1!\,n_2!\,\cdots}$.

Letters: M×1, I×4, S×4, P×2. Total 11 letters.

Letters: S×3, T×3, A×1, I×2, C×1. Total 10 letters.

Fix one S in position 1. The remaining 9 letters are: S×2, T×3, A×1, I×2, C×1.

All three results are within the same order of magnitude — a direct comparison is readable.

In a circular arrangement of $n$ people, rotations are equivalent, so we fix one person and arrange the remaining $n-1$, giving $(n-1)!$ arrangements.

Fix one person; arrange the other 7 freely.

Treat the two people as a single unit → 7 units in a circle, giving $(7-1)! = 720$ arrangements. The pair can sit in 2! = 2 internal orders.

Fix person A at one seat. In an 8-seat table there is exactly one seat directly opposite; person B must occupy it (1 way). The remaining 6 people fill the other 6 seats freely.

If we also count A and B swapping positions (B fixed, A opposite), that doubles the count: $2 \times 6! = 1440$. Whether we get 720 or 1440 depends on whether the two individuals are distinguishable in the "opposite" relation — since they are distinct people, fixing either one gives $6!= 720$; however, noting that fixing A or fixing B gives the same set of arrangements, the answer is $6! = 720$. We will state the careful derivation: fix A, B is forced to the opposite seat (no additional factor), arrange the remaining 6 people: $6! = 720$.

All three results are of the same order of magnitude — the adjacency and opposite constraints reduce the unrestricted count by similar factors.

Order does not matter, no repetition — combination.

Choose any 4 of 12 members.

Fix that person; choose 3 of the remaining 11.

Complement: subtract committees that include neither of the two specific students (choose 4 from the other 10).

Choose 2 women from the 5 women AND 2 men from the 7 men — multiply by the product rule.

All four committee counts sit in the same range, making a direct bar comparison straightforward.

There are $\binom{52}{5} = 2{,}598{,}960$ equally likely hands.

Choose 2 hearts from the 13 hearts, and 3 non-hearts from the 39 non-hearts.

Complement: subtract hands with no hearts (all 5 drawn from 39 non-hearts).

Remove all 12 face cards (J, Q, K in each of 4 suits), leaving 40 cards. Choose all 5 from those 40.

All three hand counts are in the hundreds-of-thousands range and compare cleanly on one chart.

Order matters, no repetition, selecting $k$ of $n$ — $k$-permutation: $P(n,k)=\frac{n!}{(n-k)!}$.

Three distinct medals assigned to three distinct runners — order matters.

Place 4 digits in order from 9 choices without repetition.

A number is divisible by 5 iff its last digit is 0 or 5. Since digits are from 1–9, the last digit must be 5 (only option). Fix last digit = 5, then choose and order 3 of the remaining 8 digits for the first three positions.

The three results span different magnitudes but are all in the low thousands, so a single chart remains readable.

Each position independently chooses from 10 digits with repetition → sequence with repetition. "All distinct" uses a $k$-permutation.

10 choices per position, 5 positions.

First digit: 10 choices; second: 9; third: 8; fourth: 7; fifth: 6.

Complement: total minus all-different.

The three PIN counts partition the total — all-different and at-least-one-repeat sum exactly to the unrestricted total.

First digit: 1–9 (9 choices). Remaining 4 digits: 0–9 (10 choices each).

Last digit must be even (0,2,4,6,8 → 5 choices). First digit: 1–9 (9 choices). Middle three digits: 10 choices each.

First digit: 9 choices (1–9). Second digit: 9 choices (0–9 except first). Third: 8. Fourth: 7. Fifth: 6.

Total 5-digit numbers minus those with no repeated digits.

All four counts are in the tens-of-thousands range — a single chart shows how each restriction trims the total.

Order irrelevant, no replacement — combination.

Choose 2 red balls from 5 red, and 1 non-red ball from the 7 non-red balls.

Order matters, no replacement — 3-permutation.

Choose and order 2 red balls from 5: $P(5,2)=20$ ways. Choose 1 non-red from 7: 7 ways. The non-red ball can occupy any of the 3 positions: $\binom{3}{1}=3$ ways.

Unordered and ordered draws pair naturally — the ordered counts are exactly $3!$ times larger when comparing total draws.

Permutation with repeated elements: 11 positions, groups of size 4, 4, 3.

Once every ball is distinguishable, this is a full permutation of 11 distinct objects.

The ratio $\frac{11!}{11550}=4!\,4!\,3!=3456$ is exactly the number of ways to label balls within each color group — confirming that labeling multiplies the count by the product of within-group factorials.

Possible color triples $(r,b,g)$ with $r+b+g=3$, $r \leq 4$, $b \leq 4$, $g \leq 3$: list by $r=0,1,2,3$. Weighted count using the combination formula gives $\binom{11}{3}=165$ equally-likely labeled draws, but distinct color compositions number only 10 unweighted outcome types.

Order matters, no replacement, 12 distinct balls.

The fully-labeled arrangement ($11! = 39{,}916{,}800$) is astronomically larger than the other two results — it is omitted from the chart so the remaining bars remain readable.

5 choices per letter position (repetition ok), 10 choices per digit position.

Choose and order 2 distinct letters from 5; digits still repeat freely.

Order 2 of 5 letters; order 3 of 10 digits.

Three distinct medals assigned to three distinct runners — 3-permutation.

Assign 2 of the 3 medal positions to the two specific runners: $P(3,2)=6$ ordered assignments. The third medal goes to any of the remaining 10 runners.

Choose any 4 of 10 people.

Choose 2 women from 4 and 2 men from 6.

Complement: subtract committees with no women (all 4 from 6 men).

Circular permutation — fix one person, arrange the other 6.

Glue the pair into one unit → 6 units around a circle, times 2! internal orders.

36 choices per position independently.

Choose and order 5 of 36 symbols.

The ID sub-problems share a comparable scale; the committee sub-problems form a second coherent group — both are charted together for a concise overview.

Comparing the three committee outcomes from sub-problem (3) on one chart.