I want to make sure I did this problem correctly before I add it to the problems section

How many sets of a, b, c, d, e are possible if:
ac + ae + bd + bc + ad + be + 17 = 40
And if a, b, c, d, e are all integers greater than or equal to zero?

So subtracting 17 and factoring yields
(a+b)(c+d+e)=23
since 23 is prime, its only factors are 23 and 1
if a+b=1 then c+d+e=23
and if a+b=23 then c+d+e=1

the first case, a and b can add up to 1 in 2 ways, or 2C1 ways and c d and e can add up to 23 in 25C2 ways or 300 ways.
Therefore, there would be 2*300 or 600 ways for this to happen

the second case, a and b can add up to 23 in 24C1 or 24 ways and c d and e can add up to 1 in 3C1 ways or 3 ways.
Therefore, there would be 24*3 or 72 ways for this to happen

Adding 600+72 = 672

Can someone tell me whether this is done correctly? Thanks,

#Combinatorics

Easy Math Editor

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Comments

Yup. Totally correct. Well done!

Pi Han Goh - 3 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$