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Orders across exchanges

CombinatoricsMedium

A trader sends $10$ identical orders to $4$ distinct exchanges, with at least $1$ and at most $4$ orders on each exchange. How many allocations are possible?

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Hint

Remove the minimum first, then subtract allocations that break the upper cap.

Solution

Subtract $1$ from each exchange: $y 1 + y 2 + y 3 + y 4 = 6$ with $0 \leq y i \leq 3$ .
Without the upper cap there are $(3 9) = 84$ solutions.
A violation has $y i \geq 4$ ; setting $4$ aside leaves $2$ to place, giving $(3 5) = 10$ , and $4$ exchanges give $40$ .
Violations on more than one exchange at once are impossible because $8 > 6$ .
Subtract: $84 - 40 = 44$ .

Answer: $44$

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