Play with the numbers!

Probability Level 3

a + b + c + d + e = 120 \large a + b + c + d + e = 120

Let a , b , c , d , e a, b, c, d, e be positive integers which satisfy the equation above such that a ≥ 1 a \geq 1 , b ≥ 2 b \geq 2 , c ≥ 3 c \geq 3 , d ≥ 4 d \geq 4 and e ≥ 5 e \geq 5 . Then find the number of possible ordered 5-tuples ( a , b , c , d , e ) (a,b,c,d,e) .


The answer is 5563251.

Ashish Menon by Ashish Menon , 20, India

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1 solution

Ashish Menon
Jul 25, 2016

Let a = a ′ + 1 a = a' + 1 , so a ′ ≥ 0 a' \geq 0 .
Let b = b ′ + 2 b = b' + 2 , so b ′ ≥ 0 b' \geq 0 .
Let c = c ′ + 3 c = c' + 3 , so c ′ ≥ 0 c' \geq 0 .
Let d = d ′ + 4 d = d' + 4 , so d ′ ≥ 0 d' \geq 0 .
Let e = e ′ + 5 e = e' + 5 , so e ′ ≥ 0 e' \geq 0 .


So, a ′ + 1 + b ′ + 2 + c ′ + 3 + d ′ + 4 + e ′ + 5 = 120 a ′ + b ′ + c ′ + d ′ + e ′ = 105 a' + 1 + b' + 2 + c' + 3 + d' + 4 + e' + 5 = 120\\ a' + b' + c' + d' + e' = 105

Now, the number of ways of choosing ordered 5-tuples ( a ′ , b ′ , c ′ , d ′ , e ′ ) (a', b', c', d', e') is simply 105 + 5 − 1 C 5 − 1 = 109 C 4 = 5563251 ^{105+5-1}C_{5-1}\\ = ^{109}C_4\\ = \color{#3D99F6}{\boxed{5563251}}

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Can you plz explain solution thx

Ashish Sacheti - 4 years, 10 months ago

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