Adding Gigantic Numbers!

α + β + γ = 10 10 ! α ! β ! γ ! \large \sum_{\alpha + \beta+ \gamma = 10} \dfrac{10!}{\alpha! \; \beta! \; \gamma! }

Let α , β , γ \alpha, \beta, \gamma be non-negative integers. Find the value of the summation above.

2 10 + 3 10 2^{10} + 3^{10} 3 10 3^{10} 3 12 3^{12} 3 11 3^{11}

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4 solutions

Ivan Koswara
Mar 5, 2016

There are 10 objects. You color α \alpha of them red, β \beta of them green, and γ \gamma of them blue; there are 10 ! α ! β ! γ ! \frac{10!}{\alpha!\beta!\gamma!} ways to do this. Summing over all α , β , γ \alpha, \beta, \gamma whose sum is 10 gives all possible colorings of the 10 objects with red, green, and blue. But we know that the number of such colorings is just 3 10 \boxed{3^{10}} , since each object can be colored independently in three ways.

Yash Dev Lamba
Mar 4, 2016

Consider expanison of ( x + y + z ) 10 (x+y+z)^{10} .

α + β + γ = 10 10 ! α ! β ! γ ! \sum_{\alpha+\beta+\gamma=10}\frac{10!}{\alpha!\beta!\gamma!} is equal to o the sum of the coefficients in the expansion , which we get by putiing x = y = z = 1 x=y=z=1

Therefore, α + β + γ = 10 10 ! α ! β ! γ ! \displaystyle \sum_{\alpha+\beta+\gamma=10}\frac{10!}{\alpha!\beta!\gamma!} = = 3 10 \boxed{3^{10}} .

I think we are missing something.Alfa ,bete,gamma are given as positive.The answer 3^10 is true when they are non negative(including 0)

firoz khan - 5 years, 3 months ago

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I have edited it .. THNX

Yash Dev Lamba - 5 years, 3 months ago

Same approach

Rishabh Deep Singh - 5 years, 3 months ago
Andreas Wendler
Mar 4, 2016

It is simply the sum of the coefficients of a polynomial distribution that is equal 3 n 3^{n} for an n times repetition of a test having 3 distinct final states each time. Compare this please with the binomial distribution, there we get 2 n 2^{n} .

Paulo Filho
Mar 5, 2016

The problem needs to be fixed: the solution only works if the numbers are non-negative (including 0 0 ). If the numbers are positive, the answer is 55980 55980 .

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