Difference between revisions of "2021 Fall AMC 10A Problems/Problem 21"
Arcticturn (talk | contribs) (→Solution 3 (Simple)) |
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Multinomial Numbers)) |
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For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable. | For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable. | ||
− | Let <math>d</math> be the number of ways to distribute <math>20</math> balls | + | Let <math>d</math> be the number of ways to distribute <math>20</math> balls into <math>5</math> bins. We have |
<cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.</cmath> Therefore, the answer is <cmath>\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.</cmath> | <cmath>p=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{d} \text{ and } q=\frac{\binom{20}{4,4,4,4,4}}{d}.</cmath> Therefore, the answer is <cmath>\frac pq=\frac{5\cdot4\cdot\binom{20}{3,5,4,4,4}}{\binom{20}{4,4,4,4,4}}=\frac{5\cdot4\cdot\frac{20!}{3!5!4!4!4!}}{\frac{20!}{4!4!4!4!4!}}=\frac{5\cdot4\cdot(4!4!4!4!4!)}{3!5!4!4!4!}=\frac{5\cdot4\cdot4}{5}=\boxed{\textbf{(E)}\ 16}.</cmath> | ||
Revision as of 20:00, 22 November 2021
Problem
Each of the balls is tossed independently and at random into one of the
bins. Let
be the probability that some bin ends up with
balls, another with
balls, and the other three with
balls each. Let
be the probability that every bin ends up with
balls. What is
?
Solution 1 (Multinomial Numbers)
For simplicity purposes, we assume that the balls are indistinguishable and the bins are distinguishable.
Let be the number of ways to distribute
balls into
bins. We have
Therefore, the answer is
Remark
By the stars and bars argument, we get
~MRENTHUSIASM
Solution 2 (Simple)
See Also
2021 Fall AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.