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Recall that a partition of a positive integer \(n\) means a way of writing \(n\) as the sum of some positive integers, where the order of the parts does not matter. For example, there are five partitions of \(4\) :
\(\qquad 3+1\qquad 2+2\qquad 2+1+1\qquad 1+1+1+1\)
How many partitions of 12 are there that have at least four parts, such that the largest, second-largest, third-largest, and fourth-largest parts are respectively greater than or equal to 4,3,2,1?

(The partition 12 = 4 + 4 + 2 +2  is one such partition.)

Guest Jul 24, 2018
 #1
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+1

Partitions of 12 =P(12) =77

COUNT THEM!

 

12 = 12
11 + 1 = 12
10 + 2 = 12
10 + 1 + 1 = 12
9 + 3 = 12
9 + 2 + 1 = 12
9 + 1 + 1 + 1 = 12
8 + 4 = 12
8 + 3 + 1 = 12
8 + 2 + 2 = 12
8 + 2 + 1 + 1 = 12
8 + 1 + 1 + 1 + 1 = 12
7 + 5 = 12
7 + 4 + 1 = 12
7 + 3 + 2 = 12
7 + 3 + 1 + 1 = 12
7 + 2 + 2 + 1 = 12
7 + 2 + 1 + 1 + 1 = 12
7 + 1 + 1 + 1 + 1 + 1 = 12
6 + 6 = 12
6 + 5 + 1 = 12
6 + 4 + 2 = 12
6 + 4 + 1 + 1 = 12
6 + 3 + 3 = 12
6 + 3 + 2 + 1 = 12
6 + 3 + 1 + 1 + 1 = 12
6 + 2 + 2 + 2 = 12
6 + 2 + 2 + 1 + 1 = 12
6 + 2 + 1 + 1 + 1 + 1 = 12
6 + 1 + 1 + 1 + 1 + 1 + 1 = 12
5 + 5 + 2 = 12
5 + 5 + 1 + 1 = 12
5 + 4 + 3 = 12
5 + 4 + 2 + 1 = 12
5 + 4 + 1 + 1 + 1 = 12
5 + 3 + 3 + 1 = 12
5 + 3 + 2 + 2 = 12
5 + 3 + 2 + 1 + 1 = 12
5 + 3 + 1 + 1 + 1 + 1 = 12
5 + 2 + 2 + 2 + 1 = 12
5 + 2 + 2 + 1 + 1 + 1 = 12
5 + 2 + 1 + 1 + 1 + 1 + 1 = 12
5 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
4 + 4 + 4 = 12
4 + 4 + 3 + 1 = 12
4 + 4 + 2 + 2 = 12
4 + 4 + 2 + 1 + 1 = 12
4 + 4 + 1 + 1 + 1 + 1 = 12
4 + 3 + 3 + 2 = 12
4 + 3 + 3 + 1 + 1 = 12
4 + 3 + 2 + 2 + 1 = 12
4 + 3 + 2 + 1 + 1 + 1 = 12
4 + 3 + 1 + 1 + 1 + 1 + 1 = 12
4 + 2 + 2 + 2 + 2 = 12
4 + 2 + 2 + 2 + 1 + 1 = 12
4 + 2 + 2 + 1 + 1 + 1 + 1 = 12
4 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 12
4 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
3 + 3 + 3 + 3 = 12
3 + 3 + 3 + 2 + 1 = 12
3 + 3 + 3 + 1 + 1 + 1 = 12
3 + 3 + 2 + 2 + 2 = 12
3 + 3 + 2 + 2 + 1 + 1 = 12
3 + 3 + 2 + 1 + 1 + 1 + 1 = 12
3 + 3 + 1 + 1 + 1 + 1 + 1 + 1 = 12
3 + 2 + 2 + 2 + 2 + 1 = 12
3 + 2 + 2 + 2 + 1 + 1 + 1 = 12
3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 = 12
3 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
3 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
2 + 2 + 2 + 2 + 2 + 2 = 12
2 + 2 + 2 + 2 + 2 + 1 + 1 = 12
2 + 2 + 2 + 2 + 1 + 1 + 1 + 1 = 12
2 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 12
2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
2 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12
1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 12

Guest Jul 24, 2018
 #2
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0

I got it, and you are wrong anyways. I used Ferrer's diagrams and got \(\binom{5}{2} + 1 + 1 = \boxed{12}\)

Guest Jul 24, 2018
 #3
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This answer is wrong too.

 

EDIT: ignore my idiotic reply, i read the question wrong.

Guest Jul 24, 2018
edited by Guest  Jul 24, 2018
 #4
avatar+20025 
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deleted

heureka  Jul 24, 2018
edited by heureka  Jul 24, 2018
 #5
avatar+90023 
+1

There are 77 partitions of 12....

 

See WolframAlpha's answer, here :    https://www.wolframalpha.com/input/?i=partitions+of+12

 

 

cool cool cool

CPhill  Jul 24, 2018
edited by CPhill  Jul 24, 2018
 #6
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Why was this question marked with a check mark? This is not what the question is.

Guest Jul 24, 2018
 #7
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Yes, exactly, if we read the problem, we see that not all partitions of 12 work!!! Use Ferrer's diagram

Guest Jul 24, 2018
 #8
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Would this set satisfy the conditions given?

 

1-      6 + 3 + 2 + 1 = 12
2-     5 + 4 + 2 + 1 = 12
3-     5 + 3 + 3 + 1 = 12
4-     5 + 3 + 2 + 2 = 12
5-     5 + 3 + 2 + 1 + 1 = 12
6-     4 + 4 + 3 + 1 = 12
7-     4 + 4 + 2 + 2 = 12
8-     4 + 4 + 2 + 1 + 1 = 12
9-     4 + 3 + 3 + 2 = 12
10-   4 + 3 + 3 + 1 + 1 = 12
11-    4 + 3 + 2 + 2 + 1 = 12
12-   4 + 3 + 2 + 1 + 1 + 1 = 12

Guest Jul 24, 2018
 #9
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0

I got the answer, no worries

Guest Jul 24, 2018

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