Triples

Probability Level 2

Find the number of ordered triples of positive integers $(a,b,c)$ such that $a+b+c=8$ .

The answer is 21.

1 solution

Rishik Jain
Apr 7, 2016

Relevant wiki: Integer Equations - Stars and Bars

Since $a,b,c \in \mathbb{Z_+}$ , $a+b+c=8$ can be written as $1+a'+1+b'+1+c'=8$ where $a',b',c' \ge 0$ and are integers.

$\Rightarrow a'+b'+c'=5$ By Stars and Bars , the equation has $\dbinom{7}{2} = \boxed{\boxed{21}}$ solutions.

Moderator note:

Good approach using the Stars and Bars technique.

While it is often used for "non-negative integer solutions", we can often make slight tweaks to obtain what we want.

By the WIKI this would be 10ch8 after creating a bijection using 0's. How is it 7ch2. What are a' b' and c'?

Samuel Ryan - 5 years, 2 months ago

10C2 applies to the situation if a, b, c were greater than or equal to zero. The question given here says that a, b, c are positive i.e. greater than zero , thus not necessitating the equality to zero . So, as @Rishik Jain has solved, the equation is converted to a form concerning a', b', c' , where a', b', c' are each greater than or equal to zero . Now applying the bijection creation using 0's, we get 7C2.

Hope that clarifies your doubt! :-)

Aniruddha Bhattacharjee - 5 years, 2 months ago

The question asks for ordered triplets: (1, 1, 6) and (6, 1, 1) are the same solution then. Leaving (1, 1, 6), (1, 2, 5), (1, 3, 4), (2, 2, 4), (2, 3, 3)? (If you allow (1, 6, 1) etc... There are indeed 21 solutions.)

Benjamin Katz-Crowther - 5 years, 2 months ago

'Ordered' triples convey the fact that the order does matter. For example, if ordered triplets are asked then (2,3,5) is different from (3,2,5).

Rishik Jain - 5 years, 2 months ago

Very much so!

Aniruddha Bhattacharjee - 5 years, 2 months ago

@Aniruddha Bhattacharjee – Aaaah my bad. I thought it was 'a triplet of numbers in order'

Benjamin Katz-Crowther - 5 years, 2 months ago

Did the same

Aditya Kumar - 5 years ago

This is another case of stars and bars, used for positive solutions

Yuvraj Pokharna - 3 years, 7 months ago

shouldnt it be 8c3

rishabh verma - 1 month, 2 weeks ago

I got it after a little thinking, the key here is 'positive integers'. So you can't put two consecutive bars in between the stars and also you can't put any bar at the beginning. You have to put the bars in the empty spaces between stars. There are 7 empty spaces between 8 stars. You can put 2 bars (will act as 2 plus signs) in those spaces in 7C2 ways.

Samit Khan - 3 weeks, 1 day ago

Shouldn't the last line have 7C2 and not 10C2?

Aniruddha Bhattacharjee - 5 years, 2 months ago

Yes. Fixed, thanks.

Rishik Jain - 5 years, 2 months ago

Triples

The answer is 21.

1 solution

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