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Contracts with a minimum

CombinatoricsMedium

A trader buys $12$ identical contracts spread over $4$ different strikes, with at least $2$ contracts at each strike. How many allocations are possible?

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Hint

Place the required minimum first, then distribute the rest freely.

Solution

Give $2$ contracts to each strike first, which uses $8$ contracts.
The remaining $4$ contracts are distributed freely among $4$ strikes.
By stars and bars the count is $(3 4 + 3) = (3 7)$ .
Therefore the number of allocations is $35$ .

Answer: $35$

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