That condition does it all!
Let there be positive integers n 1 , n 2 , n 3 , n 4 , n 5 such that n 1 > n 2 > n 3 > n 4 > n 5 and n 1 + n 2 + n 3 + n 4 + n 5 = 2 0 . How many sets of solution ( n 1 , n 2 , n 3 , n 4 , n 5 ) are there?
Bonus: Try the same problem for positive integers n 1 , n 2 , n 3 , n 4 . Does this problem have more solutions or lesser than the above problem?
The answer is 7.
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
1 solution
Yeah both of your answers are wright! Its pretty awesome that lesser the number of variables, more harder it gets(more number of solutions)
Log in to reply
Agreed! Although, i would like to see a more algebraic solution. Mine seems like cheating :P
Log in to reply
I will try to post a algebraic solution
I wrote a computer program to calculate the number of possible sets.
For the original problem the answer turns out to be 7 .
And for the bonus problem the answer turns out to be 2 3 .