Difference between revisions of "2021 Fall AMC 10A Problems/Problem 21"
MRENTHUSIASM (talk | contribs) m (→Solution 1 (Multinomial Numbers)) |
Arcticturn (talk | contribs) (→Solution 2 (Simple)) |
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==Solution 2 (Simple) == | ==Solution 2 (Simple) == | ||
+ | Since both of the boxes will have <math>3</math> boxes with <math>4</math> balls in them, we can leave those out. There are <math>\binom {6}{3}</math> = <math>20</math> ways to choose where to place the <math>3</math> and the <math>5</math>. After that, there are <math>\binom {8}{3} = 56</math> ways to put the <math>3</math> and <math>5</math> balls being put into the boxes. For the <math>4,4,4,4,4</math> case, after we canceled the <math>4,4,4</math> out, we have <math>\binom {8}{4}</math> = <math>70</math> ways to put the <math>4</math> balls inside the boxes. Therefore, we have | ||
+ | |||
+ | I'm still working on this solution - PLEASE DO NOT EDIT | ||
+ | |||
+ | ~Arcticturn | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2021 Fall|ab=A|num-b=20|num-a=22}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:01, 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)
Since both of the boxes will have boxes with
balls in them, we can leave those out. There are
=
ways to choose where to place the
and the
. After that, there are
ways to put the
and
balls being put into the boxes. For the
case, after we canceled the
out, we have
=
ways to put the
balls inside the boxes. Therefore, we have
I'm still working on this solution - PLEASE DO NOT EDIT
~Arcticturn
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 |
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