# Combinations

• April 5th 2011, 10:39 AM
qwerty10
Combinations
How many solutions are there to x_1+x_2+x_3+x_4=25 where x_n are all non-negative and x_1 is at most 5?
Not sure where to start

could u replace x_4 as x_4'+10 to get x_1+x_2+x_3+x_4'=20

Then get... 23C3 from combinations formula for repetition is not allowed and order is irrelevant
• April 5th 2011, 11:25 AM
Plato
Quote:

Originally Posted by qwerty10
How many solutions are there to x_1+x_2+x_3+x_4=25 where x_n are all non-negative and x_1 is at most 5?

There are $\dbinom{19+4-1}{19}$ solutions in which $x_1\ge6$.
So now what do we do?
• April 5th 2011, 12:06 PM
qwerty10
would u get: 25+4-1C4-1 - 19+4-1C4-1?
• April 5th 2011, 12:13 PM
Plato
Quote:

Originally Posted by qwerty10
would u get: 25+4-1C4-1 - 19+4-1C4-1?

Yes you would. Note I changed the 5 to a 4 in my OP.