Combinatorics Challenge: Astronaut Assignment with Experience Requirements

What You Will Learn

• Handle "forced" members first. When some people are guaranteed a spot, settle them before anything else. They stop being a choice and become part of the setup.
• Tell "choosing" from "splitting." Selecting a crew from a pool and dividing a crew between labeled destinations are different counting stages, and this problem needs both.
• Use the multiplication principle on stages. Independent stages multiply. Mutually exclusive cases add. Knowing which applies is most of the battle.
• Cross-check with a structurally different method. Counting by crew, by casework, and by inclusion-exclusion all land on the same number, and this page shows how.
• Recognize the multinomial pattern hiding behind "assign n people to labeled groups of given sizes."

Check Your Foundation

You will be comfortable here if you can do the following without hesitation:

• Evaluate a binomial coefficient: C(n,k) = n! / (k!(n−k)!), and recognize the symmetry C(n,k) = C(n, n−k).
• Apply the multiplication principle (multiply when you make one choice and then another).
• Tell when order matters (permutations) from when it does not (combinations).

Solution: Method 1 — Pick the Crew First, Then Split It

The constraint in this problem is a gift. It tells us the 4 experts are not a decision but a fixed ingredient, so the whole problem collapses into two ordinary counting stages once we say so out loud.

Step 1 — Translate the constraint into a statement about the crew

Each rocket seats a fixed number of people, so exactly 3 + 4 = 7 astronauts travel. The 4 experts are among them. That leaves 7 − 4 = 3 remaining seats to be filled by the 8 non-experts.

Step 2 — Choose the 3 non-experts who also fly

Which 3 of the 8 non-experts get a seat? Order does not matter at this stage, since we are only deciding who is on the mission:

Each of these 56 choices gives a different 7-person mission crew, and each crew contains all 4 experts.

Step 3 — Split the 7-person crew between the two rockets

Now take any one crew. Choosing which 3 of its 7 members board the 3-seat rocket automatically determines the 4 who board the other one:

This count is the same for every crew, which is why a simple product works.

Step 4 — Multiply the stages

Each of the 56 crews can be split in 35 ways, and every (crew, split) pair is a distinct assignment:

Solution: Method 2 — Casework on How the Experts Are Split

Method 1 never asked where each expert goes, because it handled placement in the 35-way split. Here we ask that question directly, which is how many people first approach the problem. Let k be the number of experts on the 3-seat rocket. Then 4 − k experts are on the 4-seat rocket.

Step 1 — Find the possible values of k

The 3-seat rocket cannot hold more than 3 experts, so k ∈ {0, 1, 2, 3}. The 4-seat rocket can hold up to 4 experts, which only requires 4 − k ≤ 4 and is always true here.

Step 2 — Count the assignments for a fixed k

Build the assignment in four sub-stages:

• Choose which k experts ride the small rocket: C(4,k). The other experts automatically go on the large rocket.
• Fill the rest of the small rocket with 3 − k non-experts: C(8, 3−k).
• The large rocket holds 4 − k experts, so it needs 4 − (4−k) = k non-experts. Only 8 − (3−k) = 5 + k remain, so choose from those: C(5+k, k).

Step 3 — Tabulate the four cases

Step 4 — Add the cases

Different values of k can never describe the same assignment, so we add:

This agrees with Method 1. The casework is longer, but it reveals the shape of the answer. Splits with 2 experts in each rocket (1008) are the most common, which matches the intuition that balanced splits dominate.

Solution: Method 3 — Start Big and Subtract (Inclusion-Exclusion)

A third path ignores the constraint at first and then removes everything that violates it. It is more work here, but it is the route that scales when the "forbidden" conditions are awkward to build directly.

Step 1 — Count all assignments with no restriction

Pick 3 of the 12 for the small rocket, then 4 of the remaining 9 for the large one:

Step 2 — Define what we want to avoid

Let Ei be the event "expert i stays on Earth." We want assignments where none of the four events happens. Inclusion-exclusion says that number is

where the term for j counts assignments in which a specific set of j experts is left behind (the rockets are then filled from the other 12−j people).

Step 3 — Evaluate each term

Step 4 — Sum

Three structurally different calculations, one number. The intermediate terms are large and alternate in sign, which is exactly why this method is a good check and a poor first choice for this particular problem.

Verification

A counting answer cannot be "plugged back in," so we confirm it by recounting in ways that do not share the same steps.

• Three methods agree. The crew-then-split product (56 × 35), the casework sum (56 + 672 + 1008 + 224), and the inclusion-exclusion alternating sum all equal 1960.
• Cross-check as a probability. The unrestricted total is also 12! / (3! 4! 5!) = 27720 (3, 4, and 5 left behind). The fraction of those with all 4 experts aboard is 1960 / 27720 = 7/99. Independently, the probability that a random 7-person crew from 12 contains all 4 experts is C(8,3) / C(12,7) = 56 / 792 = 7/99. ✓
• Bounds. 1960 is positive and well below the unrestricted 27720, as it must be for a restricted subset.

Does 7/99 Feel Right?

About 7% of unrestricted assignments include all four experts. That fits an intuition check: each expert has a 7/12 chance of flying, and for four people in a row the chance is (7/12)(6/11)(5/10)(4/9) = 7/99. It is small, but not tiny. Seeing a rough "gut" probability land exactly on the exact count's ratio is a strong sign that the setup is right.

Mistakes That Are Easy to Make Here

If You See These Words...

• "Assign ... to ... groups of sizes ..." usually means labeled groups, which suggests a multinomial coefficient or a sequence of binomial choices.
• "Must be included" / "must go" means fix those people first and shrink the pool.
• "Two different rockets / teams / rooms" signals the groups are distinguishable, so no dividing by symmetry.
• Group sizes that do not add up to the whole pool (here 3 + 4 = 7 < 12) mean someone is left out, and the "left out" group is an implicit extra group.

The same structure appears when a school assigns students to buses with required chaperones, when a hospital assigns nurses to wards with required specialists, or when a coach picks lineups that must include certain starters.

The Pattern Behind This

Assigning n people into labeled groups of sizes a, b, c, ... (with the leftover also a group) is counted by the multinomial coefficient:

A "must include" restriction removes some people from the choice. If m people are forced to travel and T seats are filled in total out of n people, the general structure is

Push Further

Try each one first, then reveal the full solution to compare.

Frequently Asked Questions