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# I seriously need help please

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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.)

Jul 24, 2018

#1
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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

Jul 24, 2018
#2
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I got it, and you are wrong anyways. I used Ferrer's diagrams and got $$\binom{5}{2} + 1 + 1 = \boxed{12}$$

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Jul 24, 2018
#3
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Guest Jul 24, 2018
edited by Guest  Jul 24, 2018
#4
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deleted

Jul 24, 2018
edited by heureka  Jul 24, 2018
#5
+101761
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There are 77 partitions of 12....

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

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

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

Jul 24, 2018
#9
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I got the answer, no worries

Jul 24, 2018