Too Many

Probability Level 3

How many solutions are there to the equation a + b + c + d = 11 a+b+c+d = 11 if a , b , c , d ∈ Z + a,b,c,d \in \mathbb{Z}_+ ?

17 88 20 120

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A P by A P , 26, USA

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3 solutions

Since a , b , c , d ≥ 1 a,b,c,d \geq 1 ,

a + b + c + d = 11 ( a ′ + 1 ) + ( b ′ + 1 ) + ( c ′ + 1 ) + ( d ′ + 1 ) = 11 a ′ + b ′ + c ′ + d ′ = 7 \begin{aligned} a+b+c+d &=& 11\\ (a'+1)+(b'+1)+(c'+1)+(d'+1) &=& 11 \\ a' +b' + c' + d' &=& 7 \end{aligned} \

Which gives us ( 10 7 ) = 120 \displaystyle{{10 \choose 7}} = \boxed{120} sets of values that satisfy the condition

3 Helpful 0 Interesting 2 Brilliant 0 Confused

Did the exact same

Aditya Kumar - 5 years ago
Jon Sy
May 8, 2016

Hello I write computer program. And it tell me 120 120 . Thus it must be 120 \boxed{120} . Thank

0 Helpful 0 Interesting 0 Brilliant 0 Confused

wow Pr0 ok die

Percy 17 hax0r - 5 years ago

Direct application of stars and bars .

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