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Prove that sum(alpha+beta+gamma=10)^(10 ...

Prove that `sum_(alpha+beta+gamma=10)^(10 !)/(alpha!beta!gamma!)=3^(10)dot`

Text Solution

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Consider expansion of `(x+y+z)^(10)`
General term `T_(r ) = (10!)/(alpha!beta!gamma!) x^(alpha)y^(beta)z^(gamma)`. Where `alpha + beta + gamma = 10`
Now `underset(alpha+beta+gamma=10)(sum)(10!)/(alpha!beta!gamma!)=3^(10)`
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