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**+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**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

#5**+1 **

There are 77 partitions of 12....

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

CPhill Jul 24, 2018

#7**0 **

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**0 **

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