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How many ways are there to write 14 as a sum of 5 positive integers, not necessariloy distinct? Order also matters, like 1+1+1+1+10 is different than 10+1+1+1+1.

 

I think there's a formula for this, but I don't remember. Can someone please remind me? Thanks!

 Sep 13, 2020
 #1
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This is a partition problem of 14 into 5 parts = 23. Ordinarily, order does not matter in partitions. But, if it is, then just multiply by 2 and REVERSE the order = 23 x 2 = 46.

 

10 + 1 + 1 + 1 + 1 = 14
9 + 2 + 1 + 1 + 1 = 14
8 + 3 + 1 + 1 + 1 = 14
8 + 2 + 2 + 1 + 1 = 14
7 + 4 + 1 + 1 + 1 = 14
7 + 3 + 2 + 1 + 1 = 14
7 + 2 + 2 + 2 + 1 = 14
6 + 5 + 1 + 1 + 1 = 14
6 + 4 + 2 + 1 + 1 = 14
6 + 3 + 3 + 1 + 1 = 14
6 + 3 + 2 + 2 + 1 = 14
6 + 2 + 2 + 2 + 2 = 14
5 + 5 + 2 + 1 + 1 = 14
5 + 4 + 3 + 1 + 1 = 14
5 + 4 + 2 + 2 + 1 = 14
5 + 3 + 3 + 2 + 1 = 14
5 + 3 + 2 + 2 + 2 = 14
4 + 4 + 4 + 1 + 1 = 14
4 + 4 + 3 + 2 + 1 = 14
4 + 4 + 2 + 2 + 2 = 14
4 + 3 + 3 + 3 + 1 = 14
4 + 3 + 3 + 2 + 2 = 14
3 + 3 + 3 + 3 + 2 = 14

 Sep 13, 2020
 #2
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You are finding all positive integer solutions to $v+w+x+y+z=14$.

Imagine 14 balls. If you put 4 dividers in any of the in-between spaces (there are 13 in-between spaces for 14 balls right?), then you get the partition of 14 into 5 balls. 

So it's 13 choose 4 = $\frac{13!}{4!9!} = 13\cdot12\cdot11\cdot10/24=1430/2=715$

 Sep 14, 2020

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