21:I[344796,["/_next/static/chunks/0whdtqcc5h8md.js","/_next/static/chunks/1qo17jq3o72yb.js","/_next/static/chunks/00mfjk9z2ewie.js","/_next/static/chunks/3_0kqrj9zuodm.js","/_next/static/chunks/2q9x6spyz4qxv.js","/_next/static/chunks/2cxhk-z7t3ztw.js","/_next/static/chunks/35kg0tmkl08ow.js","/_next/static/chunks/09hukzrse-ioj.js","/_next/static/chunks/3-qfh2-yexhl7.js","/_next/static/chunks/0_-iwh06bu1or.js"],"default"] 22:T9f5,## Overview Four students independently choose books. The quantities `A`, `B`, `C`, and `D` are non-negative integers, and their total is capped at 10. We must count the ordered quadruples `(A, B, C, D)` of non-negative integers with `A + B + C + D <= 10`. ## Key Insight An "at most" constraint becomes an "exactly" constraint by adding a slack variable. Introduce `E >= 0` equal to the unused capacity, so `A + B + C + D + E = 10` exactly. Counting solutions of that equation in five non-negative unknowns is the classic stars-and-bars problem. ## Approach With `n = 10` identical stars distributed among `k = 5` non-negative variables, the number of solutions is `C(n + k - 1, k - 1) = C(14, 4)`. Evaluating, `C(14, 4) = (14 * 13 * 12 * 11) / 24 = 1001`. An equivalent route sums over the exact total `t`: for each `t` from 0 to 10 the count of `A + B + C + D = t` is `C(t + 3, 3)`, and `C(3,3) + C(4,3) + ... + C(13,3) = 1001` by the hockey-stick identity. ## Why This Works The brute-force script `verify_5256.py` in the work directory enumerates every quadruple directly with four nested loops over `A + B + C + D <= 10` and prints `brute force count of (A,B,C,D) >=0 with A+B+C+D <= 10 : 1001`. The same script prints `stars-and-bars C(10+4,4) = 1001` and `sum_{k=0..10} C(k+3,3) = 1001`, so all three independent routes agree. The answer is `1001`, the third option. Comparing the options: `286` is `C(13, 3)`, the count for the total being exactly 10 rather than at most 10 — it ignores every selection whose total falls short. `715` is `C(13, 4)`, what you get by writing `C(n + k - 2, k - 1)` with an off-by-one in the stars-and-bars formula. `1365` is `C(15, 4)`, the result of adding the slack variable but then also incrementing `n`, double-counting the relaxation. Only `1001` matches the enumeration. ## Complexity The closed form is a single binomial coefficient, `O(k)` arithmetic operations. The verification enumeration touches `C(14, 4) = 1001` quadruples, trivial to run. ## Common Pitfalls Reading "at most 10" as "exactly 10" and answering `286`. Forgetting that the students are distinguishable and counting unordered partitions of at most 10 into four parts instead of ordered quadruples. Mixing up which argument of the binomial gets the `+1`: for `k` non-negative variables summing to `n` the answer is `C(n + k - 1, k - 1)`, so here `k = 5` (after the slack variable) and `n = 10` give `C(14, 4)`, not `C(13, 4)` or `C(15, 4)`. Excluding zero from a student's count when the problem explicitly permits it.19:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"LearningResource\",\"learningResourceType\":\"Coding Question\",\"name\":\"Counting Book Selections by Four Students with a Total Cap\",\"description\":\"At a library, four students choose books independently. Let A, B, C, and D denote the numbers of books chosen by the four students, where each number can be zero or more. 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At a library, four students choose books independently. Let A, B, C, and D denote the numbers of books chosen by the four students, where each number can be zero or more.

If the total number of books chosen is at most 10, how many possible selections (A, B, C, D) are there?

(inferred — the source image shows the stem only, with no answer options visible, so the four options below were authored: the correct count plus distractors that encode the standard counting mistakes.)

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Counting Book Selections by Four Students with a Total Cap

Problem Statement

Question 1