Permutation $(a,b,c,d,e)$ that satisfy $a+b \le c+d+e $

Question

How many permutations of $(a,b,c,d,e) $ from $(1,2,3,4,5)$ that satisfy $$ a+b \le c+d+e \: ? $$

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My approach : Let us define $(a,b) \le (c,d,e)$ means $a+b \le c+d+e$.

A first example would be

$$(1,2) \le (3,4,5)$$, which has 2 permutations for $(1,2)$ and 6 permutations for $(3,4,5)$, so there are 12 for this solution only.

Continuing this, other examples are

$$(1,3) \le (2,4,5)$$ $$(1,4) \le (2,3,5)$$ $$(1,5) \le (2,3,4)$$ $$(2,3) \le (1,4,5)$$ $$(2,4) \le (1,3,5)$$ $$(2,5) \le (1,3,4)$$ $$(3,4) \le (1,2,5)$$

So the total number of permutations is $12 \times 8 = 96?$

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Is my calculation accurate? And are there better ways to solve the problem? Thanks.

Answer 1

We do basically have to check case by case. However, it is possible to be a bit more systematic about looking for cases and checking them, so that we're certain that we haven't missed any.

$1+2+3+4+5 = 15$, and half of $15$ is $7.5$. So as long as the two numbers we choose to make the left pair sum to less than $7.5$, the three remaining numbers must sum to more.

Now, we split into cases depending on what the lowest number in a pair is (I'm working unordered here, to make the counting easier):

• $1$: Here there are $4$ pairs, from $(1, 2)$ to $(1, 5)$, and they all work
• $2$: Now there are $3$ pairs, from $(2, 3)$ to $(2, 5)$, and they all work
• $3$: This time there are two pairs, and only one of them works: $(3, 4)$
• $4$: Only one pair, and it doesn't work
• $5$: No pairs

So all in all, we get that there are $10$ possible (unordered) pairs, and $8$ of them work. From here, to take order into account, we multiply by $12$ just as you did to get the final answer of $96$.