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.)
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
I got it, and you are wrong anyways. I used Ferrer's diagrams and got \(\binom{5}{2} + 1 + 1 = \boxed{12}\)
+128731 There are 77 partitions of 12....
See WolframAlpha's answer, here : https://www.wolframalpha.com/input/?i=partitions+of+12
Yes, exactly, if we read the problem, we see that not all partitions of 12 work!!! Use Ferrer's diagram
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
