A probability problem by أحمد الحلاق

Probability Level 3

Find the sum of the first 55 terms of this sequence

( 0 0 ) , ( 1 0 ) , ( 1 1 ) , ( 2 0 ) , ( 2 1 ) , ( 2 2 ) , ( 3 0 ) , … \dbinom 00, \dbinom 10, \dbinom11 , \dbinom20 , \dbinom21 , \dbinom 22 , \dbinom30 , \ldots

Notation : ( M N ) \dbinom MN denotes the binomial coefficient , ( M N ) = M ! N ! ( M − N ) ! \dbinom MN = \dfrac{M!}{N!(M-N)!} .


The answer is 1023.

أحمد الحلاق by أحمد الحلاق , 28, Palestine

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1 solution

Tom Engelsman
Jan 11, 2021

We have the nested sum:

Σ a = 0 9 Σ b = 0 a a ! b ! ( a − b ) ! = 1 , 023 \Sigma_{a=0}^{9} \Sigma_{b=0}^{a} \frac{a!}{b!(a-b)!} = \boxed{1,023}

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