Question 1210232: 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:
4
3 + 1
2 + 2
2 + 1 + 1
1 + 1 + 1 + 1
How many partitions of 17 are there that have at least three parts, such that the largest, second-largest, third-largest, and fourth-largest parts are respectively greater than or equal to 4, 3, 2, and 1?
The partition 17 = 7 + 4 + 3 + 2 + 1 is one such partition.)
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Solution:
Let a partition of 17 with at least three parts be $\lambda_1 \ge \lambda_2 \ge \lambda_3 \ge \dots \ge \lambda_k > 0$, where $k \ge 3$ and $\sum_{i=1}^k \lambda_i = 17$.
The conditions on the largest, second-largest, third-largest, and fourth-largest parts are:
$\lambda_1 \ge 4$
$\lambda_2 \ge 3$
$\lambda_3 \ge 2$
$\lambda_4 \ge 1$ (if $k \ge 4$)
Consider a partition of 17 into $k$ parts satisfying these conditions.
Let $\lambda_1 = a_1 + 4, \lambda_2 = a_2 + 3, \lambda_3 = a_3 + 2, \lambda_i = a_i + 1$ for $i \ge 4$, where $a_1 \ge a_2 \ge a_3 \ge a_4 \ge \dots \ge a_k \ge 0$.
Case 1: Exactly 3 parts ($k=3$)
$\lambda_1 \ge \lambda_2 \ge \lambda_3 > 0$
$\lambda_1 \ge 4, \lambda_2 \ge 3, \lambda_3 \ge 2$
Let $\lambda_1 = x+4, \lambda_2 = y+3, \lambda_3 = z+2$, where $x \ge y \ge z \ge 0$.
$(x+4) + (y+3) + (z+2) = 17 \implies x+y+z = 8$.
Partitions of 8 into 3 non-negative integers:
(8, 0, 0), (7, 1, 0), (6, 2, 0), (6, 1, 1), (5, 3, 0), (5, 2, 1), (4, 4, 0), (4, 3, 1), (4, 2, 2), (3, 3, 2).
These correspond to partitions of 17: 12+3+2, 11+4+2, 10+5+2, 10+4+3, 9+6+2, 9+5+3, 8+7+2, 8+6+3, 8+5+4, 7+6+4. (10 partitions)
Case 2: Exactly 4 parts ($k=4$)
$\lambda_1 \ge 4, \lambda_2 \ge 3, \lambda_3 \ge 2, \lambda_4 \ge 1$.
Let $\lambda_1 = x+4, \lambda_2 = y+3, \lambda_3 = z+2, \lambda_4 = w+1$, where $x \ge y \ge z \ge w \ge 0$.
$(x+4) + (y+3) + (z+2) + (w+1) = 17 \implies x+y+z+w = 7$.
Partitions of 7 into 4 non-negative integers:
(7, 0, 0, 0), (6, 1, 0, 0), (5, 2, 0, 0), (5, 1, 1, 0), (4, 3, 0, 0), (4, 2, 1, 0), (4, 1, 1, 1), (3, 3, 1, 0), (3, 2, 2, 0), (3, 2, 1, 1), (2, 2, 2, 1). (11 partitions)
Case 3: Exactly 5 parts ($k=5$)
$\lambda_1 \ge 4, \lambda_2 \ge 3, \lambda_3 \ge 2, \lambda_4 \ge 1, \lambda_5 \ge 1$.
$x+4+y+3+z+2+w+1+v+1 = 17 \implies x+y+z+w+v = 6$. (7 partitions)
Case 4: Exactly 6 parts ($k=6$)
Sum of min parts = $4+3+2+1+1+1 = 12$. $x+y+z+w+v+u = 5$. (5 partitions)
Case 5: Exactly 7 parts ($k=7$)
Sum of min parts = $4+3+2+1+1+1+1 = 13$. $x+y+z+w+v+u+t = 4$. (5 partitions)
Case 6: Exactly 8 parts ($k=8$)
Sum of min parts = $4+3+2+1+1+1+1+1 = 14$. $x+y+z+w+v+u+t+s = 3$. (3 partitions)
Case 7: Exactly 9 parts ($k=9$)
Sum of min parts = $4+3+2+1+1+1+1+1+1 = 15$. $x+...+r = 2$. (2 partitions)
Case 8: Exactly 10 parts ($k=10$)
Sum of min parts = $4+3+2+1+1+1+1+1+1+1 = 16$. $x+...+q = 1$. (1 partition)
Case 9: Exactly 11 parts ($k=11$)
Sum of min parts = $4+3+2+1+1+1+1+1+1+1+1 = 17$. $x+... = 0$. (1 partition)
Total = $10 + 11 + 7 + 5 + 5 + 3 + 2 + 1 + 1 = 45$.
Final Answer: The final answer is $\boxed{45}$
Answer by ikleyn(52776)  (Show Source):
You can put this solution on YOUR website! .
As I read this problem, my feeling is that it is illogical.
Indeed, it asks, how many partitions of 17 are there that have at least three parts.
"At least three parts" includes/allows 3 parts, but then it impose some conditions
on the fourth part (which may absent).
In Math, it is prohibited to write such a gibberish and then to sell (to position) it as a "problem".
To the author: you may write gibberish,
- - - - - - - - - - - - - - - but do not forget then to throw it into the garbage bin.
Or hire a special person (like me), who will do it for you.
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